The enthalpy of neutralisation is defined as the heat change accompanying the reaction of one mole of $H^{+}$ ions from an acid with one mole of $OH^{-}$ ions from a base to form one mole of water. For strong acids and strong bases, this value is constant at approximately $-57.3 \text{ kJ mol}^{-1}$ (given here as $57.0 \text{ kJ mol}^{-1}$).

In the reaction between $HNO_{3}$ and $KOH$ ($HNO_{3} + KOH \rightarrow KNO_{3} + H_{2}O$), both are strong electrolytes .

1. Identify the moles of reactants: We have $0.5 \text{ mol}$ of $H^{+}$ (from $HNO_{3}$) and $0.2 \text{ mol}$ of $OH^{-}$ (from $KOH$).
2. Identify the limiting reagent: According to the sources, the reactant present in the least stoichiometric amount is the limiting reagent . Here, $KOH$ ($0.2 \text{ mol}$) is the limiting reagent.
3. Calculate heat released: The amount of water formed is limited by the $KOH$, resulting in $0.2 \text{ mol}$ of $H_{2}O$. Heat released = (moles of water formed) $\times$ (heat of neutralisation per mole) Heat released = $0.2 \text{ mol} \times 57.0 \text{ kJ/mol} = 11.4 \text{ kJ}$.

Thus, the total heat evolved is $11.4 \text{ kJ}$, which matches Option B.