Use ${\Delta }_i{H}_1(\mathrm{Na})=496$ and ${\Delta }_i{H}_2(\mathrm{Mg})=1451$ $\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The enthalpy change for $\mathrm{Na}(g)+\mathrm{Mg}{}^{2+}(g)\to \mathrm{Na}{}^{+}(g)+\mathrm{Mg}{}^{+}(g)$ is

The reaction is the sum of ionising Na (+496) and the reverse of the second ionisation of Mg (-1451, because $\mathrm{Mg}{}^{2+}$ gains an electron to form $\mathrm{Mg}{}^{+}$). Net = 496 - 1451 = -955 kJ/mol, exothermic.