$mur=\large\frac{nh}{2\pi}$
$\therefore \large\frac{nh}{2\pi}$$=4.2178\times 10^{-34}$
Or $n=\large\frac{4.2178\times 10^{-34}\times 2\times 3.14}{6.625\times 10^{-34}}$
$\Rightarrow 4$
Thus $\large\frac{1}{\lambda}=$$R_H\bigg[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$
The transition spectral line for $4^{th}$ to $3^{rd}$ shell is
$\large\frac{1}{\lambda}$$=109678\big[\large\frac{1}{3^2}-\frac{1}{4^2}\big]$
$\Rightarrow \lambda=1.8\times 10^{-4}cm$
Hence (b) is the correct answer.