So we have to start by demonstrating that for the "ring potential", the eigenfunctions are given by:
$$\psi_n(x) \ = \ \frac{1}{\sqrt{L}} e^{2\pi i n x/L} \ \ \ \ -L/2 \ < \ x \ < \ L/2$$
We can start by setting up the Schrödinger equation for our region from $-L/2$ to $L/2$. The potential is $0$ for all possible positions of the particle, as the particle is travelling on the ring-shaped path,
so the Schrödinger equation gives us:
$$-\frac{\hbar^2}{2m} \ \frac{\partial^2 \psi}{\partial x^2} \ = \ E \psi \ \Rightarrow \ \frac{\partial^2 \psi}{\partial x^2} \ = \ -k^2 \psi \ \ \ \ \ \ \ \ k \ = \ \frac{\sqrt{2mE}}{\hbar}$$
Solving this, we get:
$$\psi(x) \ = \ Ae^{ikx} \ + \ Be^{-ikx}$$
Now, we are allowed to let $k$ run negative, to cover the negative solutions (note that both terms in the sum, for arbitrary coefficients, also solve the Schrödinger equation with the same eigenvalue).
Since the particle is on a ring, and its path "wraps around", the wavefunction must be continuous when it goes from the leftmost point, approaching $-L/2$, to the rightmost point, approaching $L/2$. Thus, we
require that:
$$\psi(-L/2) \ = \ \psi(L/2) \ \Rightarrow \ Ae^{ikL/2} \ = \ Ae^{-ikL/2}$$
We thus get:
$$e^{ikL} \ = \ i\sin(kL) \ + \ \cos(kL) \ = \ 1$$
$$\Rightarrow \ kL \ = \ 2\pi n \ \Rightarrow \ k \ = \ \frac{2 \pi n}{L} \ \ \ \ \ \ k \ \in \ \mathbb{Z}$$
This gives us a wavefunction of:
$$\psi(x) \ = \ Ae^{2\pi i n x/L}$$
Finally, we normalize on the region $-L/2$ to $L/2$, as the particle has probability $0$ of being found anywhere else:
$$\displaystyle\int_{-L/2}^{L/2} \ |A|^2 e^{2i \pi n x/L} e^{-2i \pi n x/L} dx \ = \ \displaystyle\int_{-L/2}^{L/2} \ |A|^2 dx \ = \ 1 \ \Rightarrow \ A \ = \ \frac{1}{\sqrt{L}}$$
This gives us a final wavefunction of:
$$\psi_{n}(x) \ = \ \frac{1}{\sqrt{L}} e^{2\pi i n x/L} \ \ \ \ \ n \ \in \ \mathbb{Z}$$
As you can see, this is doubly degenerate (we first introduced this degeneracy when we let $k$ run negative). Now, we introduce the perturbation given by:
$$H' \ = \ -V_0 e^{-x^2/a^2}$$
We now have to find the new, perturbed first-order energy correction using the given equation:
$$E^{1}_{\pm} \ = \ \frac{1}{2} \Big[ (W_{aa} \ + \ W_{bb}) \ \pm \ \sqrt{(W_{aa} \ - \ W_{bb}) \ + \ 4 |W_{ab}|^2} \Big]$$
With:
$$W_{ij} \ = \ \displaystyle\int \psi_{i}(x)^* H' \psi_{j}(x) dx$$
We can calculate each of the matrix element of $W$, with $a \ = \ n$ and $b \ = \ -n$, for the degenerate pairs:
$$W_{nn} \ = \ W_{(-n)(-n)} \ = \ -\frac{V_0}{L} \displaystyle\int_{-L/2}^{L/2} e^{-x^2/a^2} dx \ = \ -\frac{V_0}{L} \displaystyle\int_{-\infty}^{\infty} e^{-x^2/a^2} dx \ = \ -\frac{V_0 a \sqrt{\pi}}{L}$$
Notice how we extend the bounds to $\pm \infty$. This may seem slightly mind-boggling, but the question specifies that $a \ll L$. Consider the points at $x \ = \ \pm a$. This function is already small, and is still not even close
to $\pm L/2$. At $\pm 2a$, the perturbation is practically $0$, and is still far away from $\pm L/2$, this it is safe for us to consider the rest of the area under this curve past the point $\pm L/2$ practically $0$, thus we are allowed to extend
the bounds of integration to $\pm \infty$. Now, let's solve the ret of the integrals:
$$W_{n(-n)} \ = \ -\frac{V_0}{L} \displaystyle\int_{-\infty}^{\infty} e^{-4 \pi i n x/L} e^{-x^2/a^2} dx$$
$$W_{(-n)n} \ = \ -\frac{V_0}{L} \displaystyle\int_{-\infty}^{\infty} e^{4 \pi i n x/L} e^{-x^2/a^2} dx$$
Let's do a little bit of algebra now:
$$-\frac{1}{a^2} x^2 \ \pm \ 4 \pi i n x \ = \ -\frac{1}{a^2} (x^2 \ \pm \ \frac{4 \pi a^2 i n}{L} x) \ = \ -\frac{1}{a^2} (x^2 \ \pm \ \frac{4 \pi a^2 i n}{L} x \ - \ \frac{4 \pi^2 a^4 n^2}{L^2} \ + \ \frac{4 \pi^2 a^4 n^2}{L^2})$$
$$\Rightarrow \ -\frac{1}{a^2} (x^2 \ \pm \ \frac{4 \pi a^2 i n}{L} x \ - \ \frac{4 \pi^2 a^4 n^2}{L^2}) \ - \ \frac{4 \pi^2 a^2 n^2}{L^2} \ = \ -\frac{1}{a^2} (x \ \pm \ \frac{2 \pi a^2 i n}{L})^2 \ - \ \frac{4 \pi^2 a^2 n^2}{L^2}$$
So our integrals are given as:
$$W_{n(-n)} \ = \ -\frac{V_0}{L} \displaystyle\int_{-\infty}^{\infty} e^{-\frac{1}{a^2} (x \ + \ \frac{2 \pi a^2 i n}{L})^2} e^{-\frac{4 \pi^2 a^2 n^2}{L^2}} dx \ = \ -\frac{V_0 a \sqrt{\pi}}{L}$$
$$W_{(-n)n} \ = \ -\frac{V_0}{L} \displaystyle\int_{-\infty}^{\infty} e^{-\frac{1}{a^2} (x \ - \ \frac{2 \pi a^2 i n}{L})^2} e^{-\frac{4 \pi^2 a^2 n^2}{L^2}} dx \ = \ -\frac{V_0 a \sqrt{\pi}}{L}$$
Notice how we got rid of the exponential coefficient, as $a \ \ll \ L$, thus it is practically equal to one.
We are now in a position to calculate our first-order energy corrections:
$$E^{1}_{\pm} \ = \ \frac{1}{2} \Big[ -\frac{2V_0 a \sqrt{\pi}}{L} \Big] \ \pm \ \frac{2V_0 a \sqrt{\pi}}{L} \Big]$$
$$\Rightarrow \ E^{1}_{+} \ = \ 0 \ \Rightarrow \ E^{1}_{-} \ = \ -\frac{2V_0 a \sqrt{\pi}}{L}$$
Cool, so now that we're done with that, we can attempt to find our "good" states. Let's do $E^{1}_{+}$ first:
$$\displaystyle\int (w\psi_{n} \ + \ z\psi_{-n})^{*} e^{-x^2/a^2} (w\psi_{n} \ + \ z\psi_{-n}) dx \ = \ -\frac{V_0a\sqrt{\pi}}{L} (w^2 \ + \ 2wz \ + \ z^2) \ = \ -\frac{V_0a\sqrt{\pi}}{L} (w \ + \ z)^2 \ = \ 0 \ \Rightarrow \ w \ = \ -z$$
Now, we can do $E^{1}_{-}$:
$$-\frac{V_0a\sqrt{\pi}}{L} (w \ + \ z)^2 \ = \ -\frac{2V_0a\sqrt{\pi}}{L} \ \Rightarrow \ w \ + \ z \ = \ \sqrt{2}$$
Remembering the normalization condition, we have:
$$|z|^2 \ + \ |w|^2 \ = \ 1$$
For $E^{1}_{+}$ this is easy, we have $w \ = \ 1/\sqrt{2}$ and $z \ = \ -1/\sqrt{2}$. For $E^{1}_{-}$, we have:
$$z^2 \ + \ (\sqrt{2} \ - \ z)^2 \ = \ 2z^2 \ - \ 2\sqrt{2} z \ + \ 1 \ = \ 0 \ \Rightarrow \ z^2 \ - \ \sqrt{2} z \ + \ \frac{1}{2} \ = \ 0 \ \Rightarrow \ (z \ - \ \frac{1}{\sqrt{2}})^2 \ = \ 0 \ \Rightarrow \ z \ = \ \frac{1}{\sqrt{2}}$$
$$\Rightarrow \ w \ = \ \frac{1}{\sqrt{2}}$$
So our good states are given as:
$$\frac{\psi_{n}(x) \ + \ \psi_{-n}(x)}{\sqrt{2}} \ \ \ \ \ \ \text{and} \ \ \ \ \ \ \frac{\psi_{n}(x) \ - \ \psi_{-n}(x)}{\sqrt{2}}$$
We can write these wavefunctions in a more appealing form:
$$\frac{\psi_{n}(x) \ + \ \psi_{-n}(x)}{\sqrt{2}} \ = \ \frac{1}{\sqrt{2L}} e^{2 \pi i n x/L} \ + \ \frac{1}{\sqrt{2L}} e^{-2 \pi i n x/L} \ = \ \sqrt{\frac{2}{L}} \cos\Big(\frac{2 \pi n x}{L}\Big)$$
$$\frac{\psi_{n}(x) \ - \ \psi_{-n}(x)}{\sqrt{2}} \ = \ \frac{1}{\sqrt{2L}} e^{2 \pi i n x/L} \ - \ \frac{1}{\sqrt{2L}} e^{-2 \pi i n x/L} \ = \ \sqrt{\frac{2}{L}} \sin\Big(\frac{2 \pi n x}{L}\Big)$$
Note that we got rid of the overall phase on the final wavefunctions.