Step 1: pick the right colligative formula.
Dissolving a non-volatile solute makes water freeze at a lower temperature, and the size of this drop is given by $\Delta T_f = K_f \times m$, where $m$ is the molality. So our plan is to find the molality, multiply by $K_f$, and subtract the drop from water's normal freezing point of $0\,°C$.
Step 2: count the moles of ethylene glycol.
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{31}{62} = 0.5\ \text{mol} \]
Step 3: turn this into molality.
Molality needs moles of solute per kilogram of solvent, and 600 g of water is 0.600 kg:
\[ m = \frac{0.5}{0.600} = 0.833\ \text{mol kg}^{-1} \]
Step 4: find the drop and the new freezing point.
\[ \Delta T_f = 1.86 \times 0.833 = 1.55\ \text{K} \]
Subtracting this drop from the freezing point of pure water gives $0 - 1.55 = -1.55\,°C$.
\[ \boxed{T_f = -1.55\,°C\ (\approx 271.6\ \text{K})} \]