We know that, wavelength of hydrogen spectrum is given by
$$
\frac{1}{\lambda}=R\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)
$$
For minimum wavelength of Lyman series,
$$
\begin{aligned}
& n_{1}=1 \text { and } n_{2}=\infty \\
\therefore \quad & \frac{1}{\left(\lambda_{L}\right)_{\min }}=R\left(\frac{1}{(1)^{2}}-\frac{1}{\infty}\right)=R \\
\Rightarrow \quad\left(\lambda_{2}\right)_{\min }=\frac{1}{R}
\end{aligned}
$$
For minimum wavelength of Balmer series,
$n_{1}=2$ and $n_{2}=\infty$
$\therefore \quad \frac{1}{\left(\lambda_{B}\right)_{\min }}=R\left(\frac{1}{2^{2}}-\frac{1}{\infty}\right)$
$=R\left(\frac{1}{4}-0\right)=\frac{R}{4}$
$\Rightarrow \quad\left(\lambda_{B}\right)_{\min }=\frac{4}{R}$
$\therefore \quad \frac{\left(\lambda_{L}\right)_{\min }}{\left(\lambda_{B}\right)_{\min }}=\frac{\frac{1}{R}}{\frac{4}{R}}=\frac{1}{4}=0.25$