According to H-spectrum;
Rydberg formula $\Rightarrow \frac{1}{\lambda}=R_{\mathrm{H}}\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
For first line, $n_1=2, n_2=3$ (Balmer series)
$\lambda=656 \mathrm{~nm}$
$\begin{aligned} \frac{1}{656} & =R_{\mathrm{H}}\left[\frac{1}{2^2}-\frac{1}{3^2}\right] \\ & =\frac{5}{36} R_{\mathrm{H}}...(i)\end{aligned}$
For second line, $n_1=2$ and $n_2=4$
$$
\begin{aligned}
\frac{1}{\lambda} & =R_{\mathrm{H}}\left[\frac{1}{2^2}-\frac{1}{4^2}\right] \\
& =\frac{3}{16} R_{\mathrm{H}}...(ii)
\end{aligned}
$$
Dividing Eqs. (i) and (ii) we get,
$$
\begin{aligned}
\frac{\lambda}{656} & =\frac{5}{36} \times \frac{16}{3} \\
\lambda & =485.9 \mathrm{~nm}
\end{aligned}
$$

Similarly, wavelength of limiting line i.e. $n_2=\infty$ and $n_1=2$
$$
\begin{aligned}
\frac{1}{\lambda} & =R_{\mathrm{H}}\left[\frac{1}{2^2}-\frac{1}{\infty^2}\right]=\frac{R_{\mathrm{H}}}{4}=\frac{109737}{4} \\
\therefore \quad \lambda & =364.4 \mathrm{~nm}
\end{aligned}
$$
Hence, wavelength of second line is 485.9 nm and limiting line is 364.4 nm.