For $He^+,\large\frac{1}{\lambda}$$=R_Hz^2\bigg[\large\frac{1}{2^2}-\frac{1}{4^2}\bigg]$
For $H,\large\frac{1}{\lambda}$$=R_H\bigg[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$
Since $\lambda$ is same
$\therefore Z^2\bigg[\large\frac{1}{2^2}-\frac{1}{4^2}\bigg]=\bigg[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$
$z=2$
$\bigg[\large\frac{1}{1^2}-\frac{1}{2^2}\bigg]=\bigg[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$
$n_1=1$ and $n_2=2$
Hence (d) is the correct answer.