Answer
$10.23\;nm^{-1}$
Work Step by Step
The radial probability density $P(r)$ at radius $r$ for the hydrogen atom in its ground state is given by
$P(r)=\frac{4}{a^3}r^2e^{-2r/a}$
Now, at $r=a$,
$P(a)=\frac{4}{a^3}a^2e^{-2a/a}=\frac{4}{a}e^{-2}$
Putting known values
$P(a)=\frac{4}{5.29\times 10^{-2} nm}e^{-2}=10.23\;nm^{-1}$
