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If the root mean square velocity of hydrogen molecule at a given temperature and pressure is $$2 \text{ km s}^{-1}$$, the root mean square velocity of oxygen at the same condition in $$\text{km s}^{-1}$$ is:
For gases at same temperature,
$$v_{rms}=\sqrt{\frac{3RT}{M}}$$
So
$$v_{rms}\propto\frac{1}{\sqrt{M}}$$
Therefore
$$\frac{v_{H_2}}{v_{O_2}}=\sqrt{\frac{M_{O_2}}{M_{H_2}}}$$Molar masses:
$$M_{H_2}=2$$
$$M_{O_2}=32$$
Given
$$v_{H_2}=2\ \text{km s}^{-1}$$
So
2vO2=322\frac{2}{v_{O_2}}
=
\sqrt{\frac{32}{2}}vO22=232
$$=\sqrt{16}=4$$
Thus
$$v_{O_2}=\frac{2}{4}$$
$$=0.5\ \text{km s}^{-1}$$
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