The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths ${\lambda }_1/{\lambda }_2$ of the photons emitted in this process is

Wavelength $\lambda$ of emitted photon as an electron transits from an initial energy level $n_i$ to some final energy level $n_f$ is given by Balmer’s formula, $\frac{1}{\lambda }=R(\frac{1}{n_f^2}-\frac{1}{n_i^2})$

Where, R = Rydberg constant. 
In transition from n = 4 to n = 3, we have $\frac{1}{{\lambda }_1}=R(\frac{1}{3^2}-\frac{1}{4^2})=R\left(\frac{7}{9\times 16}\right)\mathrm{......}\left(i\right)$

In transistor from n = 3 to n = 2 we have $=R\left(\frac{5}{9\times 4}\right)\mathrm{......}\left(ii\right)$

So, from eq. (i) and (ii) the ratio of $\frac{{\lambda }_1}{{\lambda }_2}$ is $\frac{{\lambda }_1}{{\lambda }_2}=\frac{\left(\frac{9\times 16}{7R}\right)}{\left(\frac{9\times 4}{5R}\right)}=\frac{20}{7}$