The ratio of wavelengths of the last line of the Balmer series and...

For last Balmer series,
$\dfrac{1}{{\lambda}_{b}} = R\left[\dfrac{1}{{2}^{2}} - \dfrac{1}{{\infty}^{2}}\right]$
${\lambda}_{b} = \dfrac{4}{R}$
For last Lyman series,
$\dfrac{1}{{\lambda}_{l}} = R\left[\dfrac{1}{{1}^{2}} - \dfrac{1}{{\infty}^{2}}\right]$
${\lambda}_{l} = \dfrac{1}{R}$
$\dfrac{{\lambda}_{b}}{{\lambda}_{l}} = \dfrac{\dfrac{4}{R}}{\dfrac{1}{R}}$
$\dfrac{{\lambda}_{b}}{{\lambda}_{l}} = 4$

Hence this is the correct answer.