Answer
$5.54\;nm^{-1}$
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 $r=2a$,
$P(2a)=\frac{4}{a^3}\times4a^2e^{-2\times 2a/a}=\frac{16}{a}e^{-4}$
Putting known values
$P(2a)=\frac{16}{5.29\times 10^{-2} nm}e^{-4}=5.54\;nm^{-1}$
