The energy level of H-like atoms are given by
$E_{n} =\frac{Z^{2} R h c}{n^{2}} $
$=\frac{-13.6 Z^{2}}{n^{2}} eV$
For ionisation, the transition is from $n=1$ to $n=\infty$
$E_{\text {ionisation }} =E_{\infty}-E_{1} $
$=-13.6 Z^{2}\left(\frac{1}{\infty^{2}}-\frac{1}{1^{2}}\right)$
$=13.6 Z^{2} eV$
For hydrogen
$Z =1 $
$\therefore \left(E_{\text {ion }}\right)_{n} =13.6 \,eV\,\,\, ...(i)$
For singly ionised helium $(Z=2)$
$\left(E_{\text {ion }}\right)_{ He } =13.6 \times(2)^{2} eV $
$\therefore \frac{\left(E_{\text {ion }}\right)_{ He }}{\left(E_{\text {ion }}\right)_{ H }} =\frac{13.6 \times 2^{2}}{13.6}=4$
$\Rightarrow \left(E_{\text {ion }}\right)_{ He } =4 \times 13.6 \,eV$[from Eq. (i)]
$=54.4 eV$