If the wavelength of the first line of the Balmer series of hydrogen is 6561 Å, the wavelength of the second line of the series should be

4860 Å

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 $n_1=2 , n_2 = 3$

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

$\frac{\lambda_1}{\lambda_2} = \frac{80}{108} = \frac{20}{27} =4860Å$