Answer
$n=3$
Work Step by Step
Total energy $E_{n}=-(13.6\,eV)\frac{Z^{2}}{n^{2}}$ where $n=1,2,3,...$
For hydrogen atom, $Z=1$ and for ground state $n=1$.
Total energy of a ground-state electron in hydrogen atom is
$E_{1}=-(13.6\,eV)\frac{1^{2}}{1^{2}}=-13.6\,eV$
We need the value of $n$ for which a doubly ionized lithium atom ($Z=3$) has the same total energy as a ground-state electron in hydrogen atom.
$-(13.6\, eV)\frac{3^{2}}{n^{2}}=-13.6\,eV\implies n=3$
