1. Identify the species

   The ions are: $N^{3-},\\ O^{2-},\ \text{and}\ F^-$

   Each of these has 10 electrons, so they are isoelectronic with Ne.

2. Rule for isoelectronic species

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

   Here, the nuclear charges are:

   • For $N^{3-}$: $Z = 7$
   • For $O^{2-}$: $Z = 8$
   • For $F^{-}$: $Z = 9$

   Since all have the same number of electrons, the ion with the smallest nuclear charge holds the electron cloud least tightly and hence has the largest radius.

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

3. Match with given values

   The three radii given are $1.71, 1.40, 1.36$ Å.

   According to the order above:

   • Largest radius 4ightarrow N^{3-} = 1.71\,\text{}
   • Next 4ightarrow O^{2-} = 1.40\,\text{}
   • Smallest 4ightarrow F^- = 1.36\,\text{}

4. Check options

   • A: $1.36, 1.71, 1.40$  wrong
   • B: $1.71, 1.40, 1.36$  correct
   • C: $1.71, 1.36, 1.40$  wrong
   • D: $1.36, 1.40, 1.71$  wrong

Hence, the correct option is B.