The second line of Balmer series has wavelength $4861A^0$ . The wavelength of the first line of Balmer series is ..

For the first line in balmer series: $\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5\mathrm{R}}{36}$

For second balmer line: $\frac{1}{4861}=\mathrm{R}\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3\mathrm{R}}{16}$

Divide both equations: $\frac{\mathrm{\lambda }}{4861}=\frac{3\mathrm{R}}{16}\times \frac{36}{5\mathrm{R}}$

                                                    $\mathrm{\lambda }=4861\times \frac{27}{20}$=6563 A^o