In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $$L$$. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be:

$$\frac{1}{\lambda} = R_H Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$

$$\text{L-shell: } n = 2, \quad \text{M-shell: } n = 3, \quad \text{N-shell: } n = 4$$

$$\frac{1}{L} = R_H Z^2 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R_H Z^2 \left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5}{36} R_H Z^2$$

$$\frac{1}{L'} = R_H Z^2 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = R_H Z^2 \left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3}{16} R_H Z^2$$

$$\frac{L'}{L} = \frac{\frac{5}{36} R_H Z^2}{\frac{3}{16} R_H Z^2} = \frac{5}{36} \times \frac{16}{3} = \frac{5 \times 4}{9 \times 3} = \frac{20}{27} \implies L' = \frac{20}{27}L$$