c

For second Balmer line $n_1=2 , n_2 = 4$

$\Rightarrow \frac{1}{\lambda_1} = R ( \frac{1}{2^2} - \frac{1}{4^2}) = \frac{3R}{16}$

For first Balmer line $n_1=2 , n_2 = 3$

$\Rightarrow \frac{1}{\lambda_2} = R ( \frac{1}{2^2} - \frac{1}{3^2}) = \frac{5R}{36}  $

$\frac{\lambda_1}{\lambda_2} = \frac{80}{108} = \frac{20}{27} =4860A^0 $