This problem was really fun (at this point, basically every problem is fun)! We are asked to find
the most probable value of $r$ for the ground state of the hydrogen atom! Well, we know that the
wavefunction for the ground state of the hydrogen atom is given as:
$$\psi_{100}(r, \ \theta, \ \phi) \ = \ \frac{1}{\sqrt{\pi a^3}}e^{-r/a} \ = \ R_{10}(r)Y^{0}_{0}(\theta, \ \phi) \ = \
\frac{2}{\sqrt{a^3}} e^{-r/a} Y^{0}_{0}(\theta, \ \phi)$$
Now, it is evident that if we allow the values of $\theta$ and $\phi$ to be arbitrary, then we can say that $Y \ = \ 1$, and therefore say
that the probability of finding the particle between $r$ and $r \ + \ \text{dr}$ is given as:
$$|R_{10}|^2 \ r^2 \ \text{dr} \ = \ \frac{4r^2}{a^3} e^{-2r/a} \ \text{dr}$$
Now, we are asked to find the maximum of this function (I started calculating the expectation value, and then I realized that I was being stupid).
In order to find the most likely state, this function must be maximized. Well this is easy! All we have to do is take the derivative, and find when this function
is equal to $0$ (this will correspond to either minima or maxima), however, in the case of this function, which clearly has an exponential decay factor that
dominates the function much more than the $r^2$ factor, this point will correspond to a maxima. Anyways, let's take the derivative:
$$\frac{\text{d}|R|^2}{\text{dr}} \ = \ \frac{4}{a^3} \ \frac{\text{d}}{\text{dr}} \ r^2 e^{-2r/a} \ = \ 2r e^{-2r/a} \ - \ \frac{2r^2}{a} e^{-2r/a}$$
And let's set this to $0$ (remember that exponentials can never be $0$):
$$2r e^{-2r/a} \ - \ \frac{2r^2}{a} e^{-2r/a} \ = \ 0 \ \Rightarrow \ 2r \ - \ \frac{2r^2}{a} \ = \ 0 \ \Rightarrow \ 2r\Big(1 \ - \ \frac{r}{a} \Big) \ = \ 0$$
This means that either $r \ = \ 0$, which can't be the case (as Griffiths says in the textbook), and it doesn't make sense. This leaves the case of $r \ = \ a$,
which does make A LOT of sense! Griffiths defined $a$ as the Bohr radius, which is the "radius" at which electrons in the ground state "orbit" around the nucleus.
Of course it would make sense that the most probable $r$ of the electron is here!