To determine the mass of ethylene glycol (antifreeze) needed to lower the freezing point of water to $-24^{\circ} \mathrm{C}$, we can use the freezing point depression equation, which is a colligative property given by:

$$ \Delta T_f = i \cdot K_f \cdot m $$

where:

$\Delta T_f$ is the depression in the freezing point.
$i$ is the van't Hoff factor, which is the number of particles the solute splits into in solution. For ethylene glycol, which does not dissociate in solution, $i = 1$.
$K_f$ is the cryoscopic constant (freezing point depression constant) of water, which is given as $1.86 \mathrm{~K} ~\mathrm{kg} ~\mathrm{mol}^{-1}$.
$m$ is the molality of the solution.

However, molality is defined as the number of moles of solute per kilogram of solvent, so we first need to find the molality that corresponds to the desired freezing point depression:

$$ \Delta T_f = -24^{\circ} \mathrm{C} - (0^{\circ} \mathrm{C}) = -24^{\circ} \mathrm{C} $$

Now, we can solve for $m$:

$$ m = \frac{\Delta T_f}{i \cdot K_f} $$

Substituting the given values:

$$ m = \frac{-24^{\circ} \mathrm{C}}{1 \cdot 1.86 \mathrm{~K} ~\mathrm{kg} ~\mathrm{mol}^{-1}} $$

Note that $\mathrm{C}$ and $\mathrm{K}$ are interchangeable when calculating changes in temperature. Now, we can compute the molality:

$$ m = \frac{-24}{1.86} $$

Next, we calculate the molality of the ethylene glycol:

$$ m \approx -12.903 \mathrm{~mol} ~\mathrm{kg}^{-1} $$

Keep in mind that molality is always positive, but the negative sign indicates the direction of the temperature change. Since we're calculating the amount needed, we can consider it as 12.903 moles per kilogram of water.

Now we calculate the moles of ethylene glycol required for $18.6 \mathrm{~kg}$ of water:

$$ \text{moles of ethylene glycol} = m \cdot \text{mass of water (in kg)} $$ $$ \text{moles of ethylene glycol} = 12.903 \mathrm{~mol} ~\mathrm{kg}^{-1} \cdot 18.6 \mathrm{~kg} $$ $$ \text{moles of ethylene glycol} \approx 240.0158 \mathrm{~mol} $$

Next, we find the mass of ethylene glycol needed using its molar mass:

$$ \text{mass of ethylene glycol} = \text{moles of ethylene glycol} \cdot \text{molar mass of ethylene glycol} $$

Given the molar mass of ethylene glycol is $62 \mathrm{~g} ~\mathrm{mol}^{-1}$, we then convert it to $\mathrm{kg} ~\mathrm{mol}^{-1}$ by dividing by 1000:

$$ \text{molar mass of ethylene glycol} = 0.062 \mathrm{~kg} ~\mathrm{mol}^{-1} $$

Now we can calculate the mass:

$$ \text{mass of ethylene glycol} = 240.0158 \mathrm{~mol} \cdot 0.062 \mathrm{~kg} ~\mathrm{mol}^{-1} $$ $$ \text{mass of ethylene glycol} \approx 14.88098 \mathrm{~kg} $$

Therefore, approximately $14.881 \mathrm{~kg}$ of ethylene glycol needs to be added to $18.6 \mathrm{~kg}$ of water to lower the freezing point to $-24^{\circ} \mathrm{C}$.