${\mathrm{Al}}^{3+}$ has a lower ionic radius than ${\mathrm{Mg}}^{2+}$ ion because

Both the species are isoelectronic species, as they all contain the same number of electrons i.e,10.

For isoelectronic species, the size depends upon the effective nuclear charge. More the positive charge, the more the effective nuclear charge, and thus, the lesser the size.

${\mathrm{Al}}^{3+}$ has a lower ionic radius than ${\mathrm{Mg}}^{2+}$ ion because it has a higher nuclear charge than ${\mathrm{Mg}}^{2+}$.

Hence, option(b) is correct.