The R.M.S. speeds of the molecules of Hydrogen, Oxygen, and Carbon dioxide at the same temperature are $$v_H$$, $$v_O$$ and $$v_C$$ respectively, then:

We recall the expression for the root-mean-square (R.M.S.) speed of the molecules of an ideal gas. For one mole we write

$$v_{\text{rms}}=\sqrt{\frac{3RT}{M}}$$

where $$R$$ is the universal gas constant, $$T$$ is the absolute temperature, and $$M$$ is the molar mass of the gas.

From this formula we clearly see that at a fixed temperature $$T$$ the only variable factor is the molar mass $$M$$ that appears in the denominator under the square root. Hence,

$$v_{\text{rms}}\propto\frac{1}{\sqrt{M}}.$$

In simple words, the lighter the gas (smaller $$M$$), the larger its R.M.S. speed, and vice versa.

Now we list the molar masses of the three given gases:

$$M_{\text{H}_2}=2\;\text{g mol}^{-1},\qquad M_{\text{O}_2}=32\;\text{g mol}^{-1},\qquad M_{\text{CO}_2}=44\;\text{g mol}^{-1}.$$

The order of their molar masses is therefore

$$2 \lt 32 \lt 44.$$

Because the R.M.S. speed is inversely proportional to the square root of the molar mass, the order of the speeds will be exactly the reverse:

$$v_H \gt v_O \gt v_C.$$

This matches Option C.

Hence, the correct answer is Option C.