$\large\frac{1}{\lambda}$$=R_H\bigg[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$
$R_H=1.1\times 10^7$ for Lyman series $n_1=1$ and $n_2=4$ (Given)
$\therefore \large\frac{1}{\lambda}=$$1.1\times 10^7\bigg[\large\frac{1}{1^2}-\frac{1}{4^2}\bigg]$
$\lambda=0.9696\times 10^{-7}m$
Hence (b) is the correct answer.