The wavelength of spectral lines is given by the Rydberg Formula: $\frac{1}{\lambda }=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ For the Balmer Series, $n_1=2$.

1. First Line ($H_{\alpha }$ line): Transition from $n_2=3\to n_1=2$. $\frac{1}{656}=R_H\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=R_H\left(\frac{1}{4}-\frac{1}{9}\right)=R_H\left(\frac{5}{36}\right)$
2. Second Line ($H_{\beta }$ line): Transition from $n_2=4\to n_1=2$. $\frac{1}{{\lambda }_2}=R_H\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=R_H\left(\frac{1}{4}-\frac{1}{16}\right)=R_H\left(\frac{3}{16}\right)$ Dividing Eq. 1 by Eq. 2: $\frac{{\lambda }_2}{656}=\frac{5/36}{3/16}=\frac{5}{36}\times \frac{16}{3}=\frac{20}{27}$ ${\lambda }_2=656\times \frac{20}{27}\approx 485.9\approx 486\text{ nm}$