This problem was kind of interesting because it involved more classical mechanics than quantum. We are asked to find the allowed energies of the rigid rotor system, which
is composed of two equal masses, each with mass $m$ that are free to rotate around a point that is a fixed distance between them. The rods stay at this fixed radius because they are attached
to a massless rod of total length $a$, with one mass at each end. Let us start by considering classical rotational kinetic energy:
$$E_{K} \ = \ \frac{1}{2} I \omega^2$$
Where $I$ is the moment of inertia, and $\omega$ is the angular velocity. Let us also consider the total angular momentum of the rotating system:
$$L \ = \ I\omega$$
We can then say that:
$$E_{K} \ = \ \frac{1}{2} I \omega^2 \ = \ \frac{1}{2} \ \frac{I^2 \omega^2}{I} \ = \ \frac{L^2}{2I}$$
Now, we know that this energy is the Hamiltonian for the system, as $L^2$ is an operator. We also know that the equations $\hat{H}\psi \ = \ E\psi$ and $L^2 \psi \ = \ \hbar^2 l(l \ + \ 1)$
have simultaneuous eigenfunctions (as was previously demonstrated), so if we have some eignestate corresponding to the "rigid rotor" Hamiltonian, it will also be an eigenstate of $L^2$.
This meaqns that we have:
$$\hat{H}\psi \ = \ E\psi \ \Rightarrow \ \frac{L^2}{2I} \psi \ = \ E\psi \ \Rightarrow \ \frac{\hbar^2 l(l \ + \ 1)}{2I}\psi \ = \ E\psi$$
Now, $\frac{\hbar^2 l(l \ + \ 1)}{2I}$ is obviously just a number, so we have $E_l \ = \ \frac{\hbar^2 l(l \ + \ 1)}{2I}$. Finally, we know that:
$$I \ = \ \displaystyle\sum_{n} \ m_{n} r_{n}^2 \ = \ m_{1} \ r_{1}^2 \ + \ m_{2} \ r_{2}^2 \ = \ m\Big( \frac{a}{2} \Big)^2 \ + \ m\Big( \frac{a}{2} \Big)^2 \ = \ \frac{1}{2} \ ma^2$$
And so, plugging this in to our previously obtained equation, we have:
$$E_l \ = \ \frac{\hbar^2 l(l \ + \ 1)}{2 \Big( \frac{1}{2} \ ma^2 \Big)} \ = \ \frac{\hbar^2 l(l \ + \ 1)}{ma^2}$$
Now, since the eigenstates correspond to the Hamiltonian (and therefore are solutions to the Schrodinger equation), Griffiths discusses how the obtained values of $l$ are positive integers. We
change the variable from $l$ to $n$ (just for convention), and we let $n \ = \ 0, \ 1, \ 2, \ ...$. We then have:
$$E_n \ = \ \frac{\hbar^2 n(n \ + \ 1)}{ma^2} \ \ \ \ \ \ n \ \in \ \mathbb{Z}^+ \ \cup \ \{0\}$$
The second part of this question is pretty easy (it just follows from previous knowledge in the textbook). As has already been proven, the normalized eigenstates of this system are simply the spherical harmonics,
but where we simply set $l \ = \ n$. It then follows that the system is $(2n \ + \ 1)$-fold degenerate, due to the fact that the value opf $m$ ranges from $-n$ to $n$, including $0$.