Using Rydberg formula,

$\frac{1}{\lambda }=R\left(\frac{1}{n_f^2}-\frac{{\displaystyle 1}}{{\displaystyle n_i^2}}\right)$, where, $R$ is Rydberg constant and $n$ is quantum number.

For first line of Lyman series

$\frac{1}{\lambda }=R\left(1-\frac{1}{4}\right)=R\left(\frac{3}{4}\right)$

$\Rightarrow \lambda =\frac{4}{3R} ...\left(1\right)$

For $3^{\mathrm{rd} }$ line of Paschen series

$\frac{1}{{\lambda }_3}=\mathrm{R}\left(\frac{1}{3^2}-\frac{1}{6^2}\right)=\frac{\mathrm{R}}{9}\times \frac{3}{4}$

For $2^{\mathrm{nd}}$ line of Balmer series

$\frac{1}{{\lambda }_2}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{R}{4}\times \frac{3}{4}$

Thus, $\mathrm{a}\lambda ={\lambda }_3-{\lambda }_2=\frac{12}{R}-\frac{16}{3R}=\frac{20}{3R}$

Putting equation $\left(1\right)$

$\mathrm{a}\left(\frac{4}{3R}\right)=\frac{20}{3R}\Rightarrow \mathrm{a}=5$