Total energy liberated during transition of electron from $n^{th}$ shell to first excited state
(i.e $2^{nd}$ shell) = 10.20+17.0 = 27.20 eV
= $27.20\times1.602\times10^{-12}$ erg
$\large\frac{hc}{\lambda}$= $R_H \times Z^2hc[\large\frac{1}{2^2}-\large\frac{1}{n^2}]$
$27.20\times1.602\times10^{-12}$ = $R_H \times Z^2hc[\large\frac{1}{2^2}-\large\frac{1}{n^2}]$..... (1)
Similarly total energy liberated during transition of electrons from $n^{th}$ shell to second excited state
(i.e $3^{rd}$ shell ) = 4.25+5.95 = 10.20 eV
= $10.20\times1.602\times10^{-12}$ erg
$\therefore10.20\times1.602\times10^{-12}$ = $R_H \times Z^2hc[\large\frac{1}{3^2}-\large\frac{1}{n^2}]$..... (2)
Dividing Eq (1) by Eq (2)
We get n = 6
On substituting the value of n in equation (1) or (2) we get z=3
Hence answer is (a)