If λmax is 6563 Å, then wave length of second line for Balmer series…

For Balmer series $\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$ where n = 3, 4, 5

For second line n = 4

So $\frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3}{16} R \Rightarrow \lambda=\frac{16}{3 R}$