Answer
$-1.51\,eV$
Work Step by Step
Radius of $n$th orbit $r_{n}=(5.29\times10^{-11}\,m)\frac{n^{2}}{Z}$
$=4.761\times10^{-10}\,m$
$Z=1$ for hydrogen atom.
$n^{2}=\frac{4.761\times10^{-10}\,m}{5.29\times10^{-11}\,m}=9$
$\implies n=3$
$E_{3}=(-13.6\,eV)\frac{Z^{2}}{n^{2}}=\frac{-13.6\,eV}{9}=-1.51\,eV$
