If 3.365 g of ethanol (l) is burnt completely in a bomb calorimeter...

Molar mass of ethanol ($C_2H_5OH$) $= 2(12) + 6(1) + 16 = 46$ g mol$^{-1}$

Number of moles of ethanol burnt, $n = \dfrac{3.365}{46} = 0.07315$ mol

Since the combustion occurs in a bomb calorimeter (constant volume), the heat produced corresponds to the change in internal energy ($\Delta U_c^\circ$).
$\Delta U_c^\circ = -\dfrac{99.472}{0.07315} = -1359.8$ kJ mol$^{-1}$

The balanced chemical equation for the combustion of ethanol is:
$C_2H_5OH(l) + 3O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$

Change in the number of gaseous moles, $\Delta n_g = 2 - 3 = -1$

Now, calculating the standard enthalpy of combustion ($\Delta H_c^\circ$):
$\Delta H_c^\circ = \Delta U_c^\circ + \Delta n_g RT$
$\Delta H_c^\circ = -1359.8 + (-1) \times (8.314 \times 10^{-3} \text{ kJ K}^{-1} \text{ mol}^{-1}) \times 298.15 \text{ K}$
$\Delta H_c^\circ = -1359.8 - 2.48 = -1362.28$ kJ mol$^{-1}$

The standard enthalpy of combustion can also be written in terms of enthalpies of formation:
$\Delta H_c^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$
$\Delta H_c^\circ = [2\Delta H_f^\circ(CO_2, g) + 3\Delta H_f^\circ(H_2O, l)] - [\Delta H_f^\circ(C_2H_5OH, l) + 3\Delta H_f^\circ(O_2, g)]$

Substituting the given values (note that $\Delta H_f^\circ$ of $O_2(g)$ is $0$):
$-1362.28 = [2(-393.5) + 3(-285.8)] - \Delta H_f^\circ(C_2H_5OH, l)$
$-1362.28 = [-787.0 - 857.4] - \Delta H_f^\circ(C_2H_5OH, l)$
$-1362.28 = -1644.4 - \Delta H_f^\circ(C_2H_5OH, l)$

$\Delta H_f^\circ(C_2H_5OH, l) = -1644.4 + 1362.28 = -282.12$ kJ mol$^{-1}$

The magnitude is $|\Delta H_f^\circ| = 282.12$ kJ mol$^{-1} = 2.8212 \times 10^2$ kJ mol$^{-1}$

Rounding to the nearest integer, we get $3$.

Answer: $3$