Answer
$291\;nm^{-3}$
Work Step by Step
The wave function for the ground state of the hydrogen atom, as obtained by solving the three-dimensional Schrödinger equation and normalizing the result, is
$\psi(r)=\frac{1}{\sqrt \pi a^{3/2}}e^{-r/a}$
Therefore, the probability density for the hydrogen atom in its ground state is given by
$\psi^2(r)=\frac{1}{ \pi a^{3}}e^{-2r/a}$
$\therefore$ The probability density for the hydrogen atom in its ground state at $r=a$ is given by
$\psi^2(a)=\frac{1}{ \pi a^{3}}e^{-2a/a}=\frac{1}{ \pi a^{3}}e^{-2}$
Putting known values, we obtain
$\psi^2(a)=\frac{1}{ \pi\times (5.29\times10^{-2}\;nm)^{3}}e^{-2}=291\;nm^{-3}$
