The ionic radii of $$O^{2-}$$, $$F^-$$, $$Na^+$$ and $$Mg^{2+}$$ are in the order:

We first observe that the ions $$O^{2-},\;F^- ,\;Na^+$$ and $$Mg^{2+}$$ all possess exactly ten electrons. This is because

$$O^{2-}: 8\text{ (protons)}+2\text{ (extra electrons)}=10\text{ e}^-$$
$$F^-: 9+1=10\text{ e}^-$$
$$Na^+: 11-1=10\text{ e}^-$$
$$Mg^{2+}: 12-2=10\text{ e}^-$$

Species having the same number of electrons are called isoelectronic. For an isoelectronic series the principal factor governing ionic radius is the magnitude of the nuclear charge $$Z$$: the larger the value of $$Z$$, the stronger the electrostatic attraction exerted on the electron cloud, and hence the smaller the ionic radius.

Let us therefore list the nuclear charges explicitly:

$$\begin{aligned} O^{2-}:&\; Z=8,\\[2pt] F^-:&\; Z=9,\\[2pt] Na^+:&\; Z=11,\\[2pt] Mg^{2+}:&\; Z=12. \end{aligned}$$

Because ionic radius $$r$$ varies inversely with $$Z$$ within an isoelectronic sequence, we write symbolically

$$r\propto\frac1Z.$$

Hence a smaller nuclear charge gives a larger radius. Sorting the ions from the smallest nuclear charge to the largest nuclear charge gives

$$8\;(O^{2-}) < 9\;(F^-) < 11\;(Na^+) < 12\;(Mg^{2+}).$$

Correspondingly, the ionic radii follow the reverse order:

$$r(O^{2-}) > r(F^-) > r(Na^+) > r(Mg^{2+}).$$

Comparing with the options, this sequence matches Option B.

Hence, the correct answer is Option B.