Balmer series wavelength is given by,

$\frac{1}{\lambda }=R\left[\frac{1}{2^2}-\frac{1}{n^2}\right]$

For the first line, transition from $n=3$, or

$\frac{1}{{\lambda }_1}=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]=\frac{5R}{36} ...\left(\mathrm{i}\right)$

For the second line, transition from $n=4$, then

$\frac{1}{{\lambda }_2}=R\left[\frac{1}{2^2}-\frac{1}{4^2}\right]=\frac{3R}{16} ...\left(\mathrm{ii}\right)$

Dividing equation $\left(\mathrm{i}\right)$ by $\left(\mathrm{ii}\right)$, we get,

$\frac{{\lambda }_2}{{\lambda }_1}=\frac{20}{27}\Rightarrow {\lambda }_2=\frac{20}{27}\times 6561=4860 \stackrel{\circ }{\mathrm{A}}$