Explanation

(c) The root mean square velocity
$v_{rms} = \sqrt{\frac{3RT}{M}}$
where $R$ is gas constant, $T\ $the temperature and $M$ the molecular weight.
Given, $v_{He} = v_{H},\ \ \ \ T_{H} = 273\ K,\ \ M_{H} = 2,\ \ \ M_{He} = 4$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{v_{H}}{v_{He}} = \sqrt{\frac{T_{H}}{T_{He}} \times \frac{M_{He}}{M_{H}}}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1 = \sqrt{\frac{273}{T_{He}} \times \frac{4}{2}}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ T_{He} = 546\ K$
In $℃$, $T_{He} = (546 - 273)℃ = 273℃$