The correct order of the ionic radii of $$\text{O}^{2-}$$, $$\text{N}^{3-}$$, $$\text{F}^-$$, $$\text{Mg}^{2+}$$, $$\text{Na}^+$$ and $$\text{Al}^{3+}$$ is:

We first note that the species $$\text{O}^{2-},\;\text{N}^{3-},\;\text{F}^- ,\;\text{Na}^+ ,\;\text{Mg}^{2+}$$ and $$\text{Al}^{3+}$$ all possess the same total number of electrons.

Counting one by one, we have:

$$\begin{aligned} \text{N}^{3-}: &\;7+3=10\text{ e}^- \\ \text{O}^{2-}: &\;8+2=10\text{ e}^- \\ \text{F}^- : &\;9+1=10\text{ e}^- \\ \text{Na}^+: &\;11-1=10\text{ e}^- \\ \text{Mg}^{2+}: &\;12-2=10\text{ e}^- \\ \text{Al}^{3+}: &\;13-3=10\text{ e}^- \end{aligned}$$

Since every ion contains $$10$$ electrons, they are termed isoelectronic.

For any series of isoelectronic ions, we use the fact that:

$$\text{Ionic radius} \propto \frac1Z$$

where $$Z$$ is the atomic number (nuclear charge). The mathematical statement above expresses that, with the electron cloud held constant, a larger positive nuclear charge pulls the same set of electrons closer, so the radius becomes smaller. Conversely, a smaller nuclear charge allows the cloud to expand, giving a larger radius.

We therefore list the atomic numbers:

$$\begin{aligned} Z(\text{N}) &= 7,\\ Z(\text{O}) &= 8,\\ Z(\text{F}) &= 9,\\ Z(\text{Na}) &= 11,\\ Z(\text{Mg}) &= 12,\\ Z(\text{Al}) &= 13. \end{aligned}$$

Now we arrange the ions from the highest $$Z$$ (smallest radius) to the lowest $$Z$$ (largest radius):

$$\text{Al}^{3+}\;(Z=13)\; < \;\text{Mg}^{2+}\;(Z=12)\; < \;\text{Na}^+\;(Z=11)\; < \;\text{F}^-\;(Z=9)\; < \;\text{O}^{2-}\;(Z=8)\; < \;\text{N}^{3-}\;(Z=7).$$

Comparing this derived order with the given alternatives, we find perfect agreement with Option C.

Hence, the correct answer is Option C.