In the redox reaction, taking place in acidic medium:
$\text{X}\ \text{MnO}_4^-\ (aq) + \text{Y}\ \text{SO}_2(g) \rightarrow \text{Mn}^{+2}(aq) + \text{HSO}_4^-\ (aq)$,
the ratio of X:Y in a stoichiometrically balanced equation will be

The given redox reaction can be balanced by splitting it into two half-reactions.

Reduction half-reaction:
$\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{+2} + 4\text{H}_2\text{O}$

Oxidation half-reaction:
$\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{HSO}_4^- + 3\text{H}^+ + 2e^-$

To equalize the number of electrons transferred, multiply the reduction half-reaction by $2$ and the oxidation half-reaction by $5$:

$2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{+2} + 8\text{H}_2\text{O}$
$5\text{SO}_2 + 10\text{H}_2\text{O} \rightarrow 5\text{HSO}_4^- + 15\text{H}^+ + 10e^-$

Adding the two balanced half-reactions, we get the overall balanced equation:
$2\text{MnO}_4^- + 5\text{SO}_2 + \text{H}^+ + 2\text{H}_2\text{O} \rightarrow 2\text{Mn}^{+2} + 5\text{HSO}_4^-$

Comparing this with the given equation $\text{X}\ \text{MnO}_4^- + \text{Y}\ \text{SO}_2 \rightarrow \dots$, we find:
$\text{X} = 2$
$\text{Y} = 5$

Therefore, the ratio of $\text{X}:\text{Y}$ is $2:5$.

Answer: $2:5$