Answer: 0.1120$^{\circ}$K
Depression of Freezing point, $\Delta\; T_f = i\;K_f\; m$, where $i$ is the van't Hoff factor, $K_f$ is the Freezing Point Depression Constant and $m$ is the molality.
Calculating $i$ for Sodium Sulfate $\rightarrow Na_2SO_4^{2-} (s) \rightarrow 2Na^{+} + SO_4^{2-} \rightarrow i = 3$
In this case $i = 3, \; K_f = 1.86$ and $m = 0.02 \rightarrow \Delta T_f = 3 \times 1.86 \times 0.02 = 0.112$