The ionic radii (in $$\mathring{A}$$) of $$N^{3-}$$, $$O^{2-}$$ and $$F^{-}$$ are respectively:

The three ions given, $$N^{3-}$$, $$O^{2-}$$ and $$F^{-}$$, all lie in the same period and are isoelectronic; that is, each of them contains the same total number of electrons.

We count the electrons one by one. Neutral nitrogen has atomic number $$Z = 7$$, oxygen has $$Z = 8$$ and fluorine has $$Z = 9$$. Adding the negative charges:

$$N^{3-}: 7 + 3 = 10 \text{ electrons}$$

$$O^{2-}: 8 + 2 = 10 \text{ electrons}$$

$$F^{-}: 9 + 1 = 10 \text{ electrons}$$

So all the ions possess $$10$$ electrons, the same configuration as the noble-gas neon. Because they contain the same number of electrons, the deciding factor for their sizes is the pull of the nucleus on that common electron cloud. To measure that pull we quote the qualitative idea of effective nuclear charge, written formally as

$$Z_{\text{eff}} = Z - S,$$

where $$Z$$ is the actual nuclear charge (atomic number) and $$S$$ is the screening constant caused by the other electrons. Within an isoelectronic series the screening is almost identical, therefore the ion with the larger $$Z$$ exerts the larger $$Z_{\text{eff}}$$ on every electron. A larger effective nuclear charge contracts the electron cloud more strongly and the ionic radius becomes smaller.

Ordering the nuclear charges we have

$$Z(N^{3-}) = 7 < Z(O^{2-}) = 8 < Z(F^{-}) = 9.$$

Because the effective pull grows from left to right, the radii do the opposite:

$$r(N^{3-}) > r(O^{2-}) > r(F^{-}).$$

Now we inspect the numerical alternatives offered. The only list that follows the descending trend just obtained is

$$1.71\,\text{\AA} \; (N^{3-}) \; > \; 1.40\,\text{\AA} \; (O^{2-}) \; > \; 1.36\,\text{\AA} \; (F^{-}).$$

These three values appear together in Option D.

Hence, the correct answer is Option D.