For Balmer Series
$\large\frac{1}{\lambda}$=$R_H[\large\frac{1}{n_1^2}-\large\frac{1}{n_2^2}]$
$\Rightarrow\large\frac {1} {4103\times10^{-8} }$= $109678[\large\frac{1}{2^2}-\large\frac{1}{(m+2)^2}]$
$\Rightarrow [\large\frac{1}{2^2}-\frac{1}{(m+2)^2}] = \frac{10^8}{4103 \times 109678}$
$\Rightarrow \large\frac{1}{(m+2)^2} = 0.02778$
$\Rightarrow\large\frac{1}{(m+2)} = 0.1667$
$\Rightarrow m+2 =5.99 \approx 6$
$\therefore m = 4$
$m^{th}$ line represents $4^{th}$ line of balmer series
Hence answer is (c)