The radius of which ion is closest to that of the ${\mathrm{Li}}^{+}$ ion?

${\mathrm{Mg}}^{2+}$ is the ion whose radius is closest to that of ${\mathrm{Li}}^{+}$.

The result of the $\left(\frac{\mathrm{Z}}{\mathrm{e}}\right)$ ratio is $(11/10)>(12/10)>(3/2)>(4/2)$; the order of the ionic radius is ${\mathrm{Na}}^{+}>{\mathrm{Mg}}^{2+}>{\mathrm{Li}}^{+}>{\mathrm{Be}}^{2+}$.

Thus, the option C is correct.