Explanation

(c) de-Broglie wavelength $\lambda = \frac{h}{mv_{rms}}$
$rms$ velocity of a gas particle at the given temperature $(T)$ is given as
$\frac{1}{2}mv_{rms}^{2} = \frac{3}{2}kT \Rightarrow v_{rms} = \sqrt{\frac{3kT}{m}} \Rightarrow mv_{rms} = \sqrt{3mk\ T}$
$\therefore\lambda = \frac{h}{mv_{rms}} = \frac{h}{\sqrt{3mkT}}$
$\Rightarrow \frac{\lambda_{H}}{\lambda_{He}} = \sqrt{\frac{m_{He}T_{He}}{m_{H}T_{H}}} = \sqrt{\frac{4(273 + 127)}{2(273 + 27)}} = \sqrt{\frac{8}{3}}$