If 917Å be the lowest wavelength of Lyman series then the lowest...

Using Rydberg's formula,

$\dfrac{1}{\lambda} = R{Z}^{2}\left(\dfrac{1}{{n}^{2}1} - \dfrac{1}{{n}^{2}2}\right)$

For the lowest wavelength in Lyman series, ${n}_{1} = 1, {n}_{2} = \infty$.

$\dfrac{1}{\lambda} = R{Z}^{2}\left(\dfrac{1}{{1}^{2}} - \dfrac{1}{{\infty}^{2}}\right) = R{Z}^{2}$

For Balmer series, ${n}_{1} = 2, {n}_{2} = \infty$

$\dfrac{1}{\lambda '} = R{Z}^{2}\left(\dfrac{1}{{2}^{2}} - \dfrac{1}{{\infty}^{2}}\right) = \dfrac{R{Z}^{2}}{4}$

So,

$\dfrac{1}{\lambda '} = \dfrac{1}{4\lambda} ⇒ \lambda ' = 4 \times 917\overset{}{Å = 3668Å}$