The solution which have lowest freezing point is

The freezing point depression is given by the formula $\Delta T_f = i \times K_f \times m$, where $i$ is the van't Hoff factor, $K_f$ is the cryoscopic constant, and $m$ is the molality. Since $K_f$ is constant for a given solvent and the molarity $M$ is given as $0.2 \mathrm{M}$ for all solutions, the freezing point depression is directly proportional to the van't Hoff factor $i$. The solution with the highest $i$ will have the lowest freezing point.

For $0.2 \mathrm{M} \mathrm{K}_2 \mathrm{SO}_4$: $\mathrm{K}_2 \mathrm{SO}_4 \rightarrow 2\mathrm{K}^{+} + \mathrm{SO}_4^{2-}$, so $i = 3$.

For $0.2 \mathrm{M} \mathrm{KCl}$: $\mathrm{KCl} \rightarrow \mathrm{K}^{+} + \mathrm{Cl}^{-}$, so $i = 2$.

For $0.2 \mathrm{M} \mathrm{NaNO}_3$: $\mathrm{NaNO}_3 \rightarrow \mathrm{Na}^{+} + \mathrm{NO}_3^{-}$, so $i = 2$.

For $0.2 \mathrm{M} \mathrm{MgSO}_4$: $\mathrm{MgSO}_4 \rightarrow \mathrm{Mg}^{2+} + \mathrm{SO}_4^{2-}$, so $i = 2$.

Comparing the values of $i$, the solution with $i = 3$ will have the largest depression in freezing point, resulting in the lowest freezing point.

Answer: $0.2 \mathrm{M} \mathrm{K}_2 \mathrm{SO}_4$