The amount of energy released when $20\mathrm{ml}$ of $0.5\mathrm{M}\mathrm{NaOH}$ are mixed with $100\mathrm{ml}$ of $0.1$$\mathrm{M} \mathrm{HCl}$ is x $\mathrm{KJ}$. The heat of neutralization (in $\mathrm{KJ}{\mathrm{mol}}^{-1}$ ) is

Milli moles (or milli equivalents) of $\mathrm{NaOH}=20\times 0.5=10\mathrm{milli} \mathrm{moles}$.

Milli moles (or milli equivalents) of $\mathrm{HCl}=100\times 0.1=10\mathrm{milli} \mathrm{moles}$.

$$\begin{gathered} \therefore \underset{-10 \\ \text{ milli equivalents }}{\mathrm{NaOH}}+\underset{ -10 \\ \text{ milli equivalents }}{\mathrm{HCl}}\to \mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O} \end{gathered}$$

Thus, the heat is released when 10 milli equivalents of $\mathrm{HCl}$ are. neutralized by 10 millimoles of $\mathrm{NaOH}$ $=\mathrm{x}\mathrm{kJ}$. But the heat of neutralization is heat released when 1 equivalent of $\mathrm{HCl}$ is neutralized by 1 equivalent of $\mathrm{NaOH}$

$\therefore {\mathrm{\Delta }}_{\text{neu }}\mathrm{H}=-\frac{\mathrm{x}}{10\times {10}^{-3}}\text{ or }-100\times \mathrm{kJ}/\mathrm{mol}$

Hence option A  is the correct answer.