The correct order of ionic size of $$N^{3-}, Na^+, F^-, Mg^{2+}$$ and $$O^{2-}$$ is :

Identify the species and their isoelectronic nature.

All the given ions are isoelectronic with 10 electrons each:

$$N^{3-}$$ (7 + 3 = 10 e$$^-$$), $$O^{2-}$$ (8 + 2 = 10 e$$^-$$), $$F^-$$ (9 + 1 = 10 e$$^-$$), $$Na^+$$ (11 - 1 = 10 e$$^-$$), $$Mg^{2+}$$ (12 - 2 = 10 e$$^-$$).

Apply the rule for isoelectronic species.

For isoelectronic species, the ionic radius decreases as the nuclear charge (atomic number) increases. This is because a higher nuclear charge pulls the same number of electrons closer to the nucleus.

Arrange in order of increasing ionic size.

Nuclear charges: $$Mg^{2+}$$ (Z=12) > $$Na^+$$ (Z=11) > $$F^-$$ (Z=9) > $$O^{2-}$$ (Z=8) > $$N^{3-}$$ (Z=7)

Therefore, ionic size increases as:

$$Mg^{2+} < Na^+ < F^- < O^{2-} < N^{3-}$$

This matches Option A.

The answer is $$\boxed{\text{Option A}}$$.