Assuming complete dissociation of the electrolytes, which of the...

The depression in freezing point is given by $\Delta T_f = i \times K_f \times m$.

Assuming molarity is approximately equal to molality for dilute aqueous solutions, $\Delta T_f \propto i \times C$, where $i$ is the van 't Hoff factor and $C$ is the concentration.

For $0.4\text{ M}$ Ethylene glycol: It is a non-electrolyte, so $i = 1$.
$i \times C = 1 \times 0.4 = 0.4$

For $0.15\text{ M CaCl}_2$: It dissociates into $3$ ions ($\text{Ca}^{2+}$ and $2\text{Cl}^-$), so $i = 3$.
$i \times C = 3 \times 0.15 = 0.45$

For $0.1\text{ M K}_4[\text{Fe}(\text{CN})_6]$: It dissociates into $5$ ions ($4\text{K}^+$ and $[\text{Fe}(\text{CN})_6]^{4-}$), so $i = 5$.
$i \times C = 5 \times 0.1 = 0.5$

For $0.2\text{ M NaBr}$: It dissociates into $2$ ions ($\text{Na}^+$ and $\text{Br}^-$), so $i = 2$.
$i \times C = 2 \times 0.2 = 0.4$

The solution with the highest value of $i \times C$ will have the maximum depression in freezing point and thus the lowest freezing temperature. This corresponds to $0.1\text{ M K}_4[\text{Fe}(\text{CN})_6]$.

Answer: $0.1\text{ M K}_4[\text{Fe}(\text{CN})_6]$