The probability that the electron is between $r$ and $r + dr$ is

$\begin{align*}\left|\frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0} \right|^2 4 \pi r^2 dr\end{align*}$
So, the probability is proportional to $f(r) = r^2 e^{-2 r/(a_0)}$ . We can find the maximum value of the probability by differentiating $f$ with respect to $r$ and setting the derivative equal to $0$ .

$\begin{align*}\frac{d f}{dr} = 2 r e^{-2 r/a_0} - \frac{2 r^2}{a_0} e^{-2 r/a_0} = 0 \Rightarrow r = a_0\end{align*}$

We can check that this is a local maximum by finding the second derivative

$\begin{align*}\frac{d^2 f}{dr^2} = 2 e^{-2 r/a_0} - \frac{4 r}{a_0} e^{-2 r/a_0} - \frac{4 r}{a_0 }e^{-2 r/a_0} + \frac{4 r^...$
If we plug in $r = a_0$ , we find that

$\begin{align*}\frac{d^2 f}{dr^2} = 2 e^{-2} - 8 e^{-2} + 4e^{-2} = -2 e^{-2} < 0 \end{align*}$
Therefore, $r = a_0$ is indeed a local maximum of the radial probability density function. Therefore, the correct answer is $a_0$ .

Solution to 2001 Problem 93