If the wavelength of the first member of Lyman series of hydrogen...

For Lyman series:

For first member, $\dfrac{1}{\lambda} = \dfrac{13 . 6{Z}^{2}}{hc}\left[\dfrac{1}{{1}^{2}} - \dfrac{1}{{2}^{2}}\right]$ ….. (i)

For second member, $\dfrac{1}{\lambda '} = \dfrac{13 . 6{Z}^{2}}{hc}\left[\dfrac{1}{{1}^{2}} - \dfrac{1}{{3}^{2}}\right]$ ….. (ii)

On dividing (i) and (ii), we get

$\lambda ' = \dfrac{27}{32}\lambda$