At a fixed temperature and pressure, 10.0 mL of oxygen, M = 32.00 g/mol, takes 38 s to effuse…

Step 1. The rule. Equal volumes have equal amounts, so the time is inversely proportional to the rate: $\dfrac{t_x}{t_{\text{O}_2}} = \sqrt{\dfrac{M_x}{M_{\text{O}_2}}}$.

Step 2. The molar mass. We want $M_x$, so $M_x = M_{\text{O}_2}\left(\dfrac{t_x}{t_{\text{O}_2}}\right)^2 = 32.00 \times \left(\dfrac{54}{38}\right)^2 = 64.6\ \text{g/mol}$.

Step 3. The gas. Sulfur dioxide has $64.07\ \text{g/mol}$. The other two are $44.01$ and $70.90\ \text{g/mol}$, which are much farther away. The gas is $\text{SO}_2$.

Step 4. The time for $\text{SO}_2$. $t = 38\ \text{s} \times \sqrt{\dfrac{64.07}{32.00}} = 53.8\ \text{s}$.