According to Bohr, the wavelength emitted when an electron jumps from $n_1$ th to $n_2$ th orbit is given as
$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
For first line of Lyman series, $n_1=1$ and $n_2=2$
$\begin{array}{llll}
& & \frac{1}{\lambda_1} & =R\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\
\therefore & \frac{1}{\lambda_1} & =R\left(1-\frac{1}{2^2}\right)=\frac{3 R}{4}
\end{array}$
For the first line of Balmer series, $n_1=2$ and $n_2=3$

$\therefore$ Dividing Eq. (ii) by Eq (i), we get
$\begin{array}{llll}
\frac{\lambda_1}{\lambda_2} =\frac{5 R}{36} \times \frac{4}{3 R} \\
\Rightarrow \quad \frac{5}{27} =9 \alpha \\
\Rightarrow \quad \alpha =\frac{5}{27 \times 9}=0.021
\end{array}$