${\mathrm{IE}}_1\text{ and }{\mathrm{IE}}_2\text{ of }\mathrm{Mg}\text{ are }178\text{ and }348 \mathrm{kcal} {\mathrm{mol}}^{-1}\text{. The energy }$$\text{ required for the}\text{ ionisation as shown is, }\mathrm{Mg}⟶{\mathrm{Mg}}^{2+}+2e^{-}\text{is }$(in same units)

$$\begin{gathered} \left(\mathrm{i}\right) {\mathrm{Mg}}_{\left(\mathrm{g}\right)}⟶{\mathrm{Mg}}_{\left(\mathrm{g}\right)}^{\oplus }+{\mathrm{e}}^{-},{\mathrm{IE}}_1=178 \mathrm{kcal} {\mathrm{mol}}^{-1} \\ \text{ (ii) }{\mathrm{Mg}}_{\left(\mathrm{g}\right)}^{\oplus }⟶{\mathrm{Mg}}_{\left(\mathrm{g}\right)}^{2+}+{\mathrm{e}}^{-}{\mathrm{IE}}_2=348 \mathrm{kcal} {\mathrm{mol}}^{-1} \end{gathered}$$

Adding (i) and (ii), we have

${\mathrm{Mg}}_{\left(\mathrm{g}\right)}⟶{{\mathrm{Mg}}^{2+}}_{\left(\mathrm{g}\right)}+2e^{-}$

$\therefore \mathrm{E}={\mathrm{IE}}_2+{\mathrm{IE}}_2=178+348=526\mathrm{kcal}$