1. Identify the species

   The ions $N^{3-}$, $O^{2-}$, and $F^-$ are isoelectronic species.

   Each has: $10 \text{ electrons}$

   • $N^{3-}$: $7+3=10$
   • $O^{2-}$: $8+2=10$
   • $F^-$: $9+1=10$

2. Use the isoelectronic trend

   In an isoelectronic series, radius decreases as nuclear charge increases.

   Nuclear charges are:

   • $N$: $Z=7$
   • $O$: $Z=8$
   • $F$: $Z=9$

   Since $N^{3-}$ has the smallest nuclear charge among these three, it will have the largest ionic radius.

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

3. Compare with the given values

   Given: $r(F^-)=1.33\,\text{\AA}, \qquad r(O^{2-})=1.40\,\text{\AA}$

   So, $r(N^{3-}) > 1.40\,\text{\AA}$

   Hence $N^{3-}$ is bigger than both $O^{2-}$ and $F^-$.

4. Compare with covalent radius of nitrogen atom

   Covalent radius of $N$ is: $0.74\,\text{\AA}$

   Since an anion is much larger than its neutral atom, $r(N^{3-}) > r(N)$

5. Check options

   • A: Smaller than $F^-$ and $N$ → false
   • B: Bigger than $O^{2-}$ and $F^-$ → true
   • C: Bigger than $F^-$ and $N$, but smaller than $O^{2-}$ → false
   • D: Smaller than $O^{2-}$ and $F^-$, but bigger than $N$ → false

Therefore, the correct option is B.