Using the boiling point elevation formula:

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

where:

$ \Delta T_b $ is the elevation in boiling point,

$ K_b $ is the ebullioscopic constant,

$ m $ is the molality.

Given:

$ K_b = 0.76 \, \text{K} \, \text{kg} \, \text{mol}^{-1} $,

$ \Delta T_b = 2^\circ\text{C} $,

mass of solute $ = 2.5 \, \text{g} $,

mass of solvent (water) $ = 100 \, \text{g} = 0.1 \, \text{kg} $.

First, find the molality $ m $:

$$ m = \frac{\Delta T_b}{K_b} = \frac{2}{0.76} = \frac{200}{76} \approx 2.63 \, \text{mol} \, \text{kg}^{-1} $$.

Now calculate the moles of solute:

Given the molality $ m $,

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

Thus,

$$ \text{moles of solute} = m \times \text{mass of solvent in kg} = 2.63 \times 0.1 = 0.263 \, \text{mol} $$.

Next, determine the molecular weight $ M $ of the solute:

$$ M = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{2.5 \, \text{g}}{0.263 \, \text{mol}} \approx 9.51 \, \text{g/mol} $$.

Using Raoult's Law for the vapour pressure of the solution:

$$ P_{\text{solution}} = P_0 \cdot (1 - \chi_{\text{solute}}) $$,

where:

$ P_0 $ is the vapour pressure of pure solvent,

$ \chi_{\text{solute}} $ is the mole fraction of the solute.

Approximate the mole fraction of the solute:

$$ \chi_{\text{solute}} = \frac{\text{moles of solute}}{\nu_{\text{solvent}} + \nu_{\text{solute}}} \approx \frac{0.263}{5.55 + 0.263} \approx \frac{0.263}{5.813} \approx 0.0452 $$,

where $\nu_{\text{solvent}}$ are the moles of water (approximately 5.55 mol for 100 g of water).

Therefore, the vapour pressure of the solution:

$$ P_{\text{solution}} = 760 \times (1 - 0.0452) \approx 760 \times 0.9548 \approx 724 \, \text{mm Hg} $$.

Thus, the vapour pressure of the solution is 724 mm of Hg, corresponding to Option A.