Answer
5
Work Step by Step
Recall: $E_{n}=-(13.6\,eV)\frac{Z^{2}}{n^{2}}$
For hydrogen atom, $Z=1$ and for the first excited state, $n=2$.
$E_{2}=-(13.6\,eV)\frac{1^{2}}{2^{2}}=-3.40\,eV$
$E_{n}=-3.40\,eV+2.86\,eV=-0.54\,eV=-(13.6\,eV)\frac{1^{2}}{n^{2}}$
$\implies n^{2}=\frac{-13.6\,eV}{-0.54\,eV}=25$
$\implies n=5$
