Determine the van't Hoff factor (i):

The electrolyte is $\mathrm{A}_3 \mathrm{~B}$. When it dissolves in water, we can assume it dissociates into 3 ions of A and 1 ion of B, meaning a total of 4 ions per formula unit if it were completely dissociated. So, $n = 4$.

The solution is 80% ionized, which means the degree of ionization ($\alpha$) is 0.80.

The van't Hoff factor ($i$) accounts for the effective number of particles in solution. For dissociation, the formula is:

$$i = 1 + \alpha(n - 1)$$

Plugging in the values:

$$i = 1 + 0.80(4 - 1)$$

$$i = 1 + 0.80(3)$$

$$i = 1 + 2.40$$

$$i = 3.40$$

Calculate the molality (m) of the solution:

Molality is defined as moles of solute per kilogram of solvent.

Based on our assumption, the moles of solute ($\mathrm{A}_3 \mathrm{~B}$) = 0.5625 mol.

The volume of water is 750 ml. Since the density of water is approximately 1 g/mL, the mass of water is 750 g, which is 0.750 kg.

Now, let's calculate the molality:

$$m = \frac{\text{moles of solute}}{\text{mass of solvent (in kg)}}$$

$$m = \frac{0.5625 \text{ mol}}{0.750 \text{ kg}}$$

$$m = 0.75 \text{ mol kg}^{-1}$$

Calculate the boiling point elevation ($\Delta T_b$):

The formula for boiling point elevation is:

$$\Delta T_b = i \cdot K_b \cdot m$$

We are given $K_b$ for water as $0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$.

$$ \Delta T_b = 3.40 \cdot 0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \cdot 0.75 \mathrm{~mol} \mathrm{~kg}^{-1}$$

$$ \Delta T_b = 1.326 \text{ K}$$

Calculate the boiling point of the solution:

The normal boiling point of pure water at 1.0 atm is $100^\circ\text{C}$, which is 373.15 K. Sometimes, for simplicity in problems, it's taken as 373 K. Let's see which one matches the options better.

$$T_{\text{b, solution}} = T_{\text{b, pure water}} + \Delta T_b$$

Using $T_{\text{b, pure water}} = 373 \text{ K}$:

$$T_{\text{b, solution}} = 373 \text{ K} + 1.326 \text{ K}$$

$$T_{\text{b, solution}} = 374.326 \text{ K}$$

This value is very close to Option A.

The final answer is $\boxed{\text{374.33 K}}$.

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