Answer
16
Work Step by Step
Energy $E_{n}=-(13.6\,eV)\frac{Z^{2}}{n^{2}}$
For the $n$th orbit of triply ionized beryllium atom ($Z=4$),
$E_{n}=-(13.6\,eV)\frac{4^{2}}{n^{2}}=-(13.6\,eV)\times\frac{16}{n^{2}}$
For the $n$th orbit of hydrogen atom ($Z=1$),
$E_{n}=-(13.6\,eV)\times\frac{1^{2}}{n^{2}}$
Ratio= $\frac{-(13.6\,eV)\times\frac{16}{n^{2}}}{-(13.6\,eV)\times\frac{1}{n^{2}}}=16$
