Ionization energy of $H{e}^{+}$ is $19.6\times {10}^{-18}Jato{m}^{-1}.$ The energy of the first stationary state of $L{i}^{2+}$ is

$E=\frac{-k\times z^2}{n^2}$

$I=\frac{k\times z^2}{n^2}$

$I\alpha \frac{z^2}{n^2}$

$\frac{I_1}{I_2}=\frac{z_1^2}{n_1^2}\times \frac{n_2^2}{z_2^2}$

$\frac{19.6\times {10}^{-18}J}{I_2}=\frac{4}{9}$

$I_2=\frac{19.6\times {10}^{-18}\times 9}{4}J$

$\begin{array}{}{I}_2=-44.1\times {10}^{-18}J\end{array}$