The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of 108.5 nm. The ground state energy of an electron of this ion will be

54.4 eV

For third line of Balmer series $n_1=2$, $n_2=5$

∴  $\frac{1}{\lambda}=R Z^2\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$ gives $Z^2=\frac{n_1^2 n_2^2}{\left(n_2^2-n_1^2\right) \lambda R}$

On putting values $Z=2$

From $E=-\frac{13.6 Z^2}{n^2}=\frac{-13.6(2)^2}{(1)^2}=-54.4 eV$