Root mean square speed $=\sqrt{\frac{3 R T}{M}}$

Given,
$$
\begin{aligned}
M_1 & =\text { molar mass of } \mathrm{H}_2=2 \\
M_2 & =\text { molar mass of } \mathrm{O}_2=32 \\
T_1 & =50 \mathrm{~K}, T_2=800 \mathrm{~K}
\end{aligned}
$$

On taking ratio of $r_1$ and $r_2$, we get
$$
\begin{aligned}
& \frac{r_1}{r_2}=\sqrt{\frac{M_2 \times T_1}{M_1 \times T_2}}=\sqrt{\frac{32}{2}} \times \sqrt{\frac{T_1}{T_2}} \\
& \frac{r_1}{r_2}=\sqrt{16} \times \sqrt{\frac{50}{800}} \text { or } \frac{r_1}{r_2}=4 \times \frac{1}{4}
\end{aligned}
$$
or
$$
\frac{r_1}{r_2}=\frac{1}{1}=1: 1
$$

Hence, option (c) is correct.