The isoelectronic set of ions is:

We recall that species are said to be isoelectronic when each of them possesses exactly the same total number of electrons.

For any neutral atom the number of electrons equals its atomic number $$Z$$. When the species is an ion, we adjust this count as follows:

For a cation having charge $$+n$$, the ion has lost $$n$$ electrons, so

$$\text{electrons in cation}=Z-n.$$

For an anion carrying charge $$-n$$, the ion has gained $$n$$ electrons, therefore

$$\text{electrons in anion}=Z+n.$$

Now we write the atomic numbers of all the elements that appear in the four options:

$$\begin{aligned} Z(\text{Li})&=3,\\ Z(\text{N}) &=7,\\ Z(\text{O}) &=8,\\ Z(\text{F}) &=9,\\ Z(\text{Na})&=11,\\ Z(\text{Mg})&=12. \end{aligned}$$

Using the above relationships we calculate the electron count for every ion mentioned in the options:

$$\begin{aligned} \text{Li}^+ &: 3-1=2,\\[2pt] \text{N}^{3-}&: 7+3=10,\\[2pt] \text{O}^{2-}&: 8+2=10,\\[2pt] \text{F}^- &: 9+1=10,\\[2pt] \text{Na}^+ &: 11-1=10,\\[2pt] \text{Mg}^{2+}&: 12-2=10. \end{aligned}$$

With these values in hand we examine each option one by one.

Option A: $$\text{F}^- (10),\; \text{Li}^+ (2),\; \text{Na}^+ (10),\; \text{Mg}^{2+} (10)$$ — the value $$2$$ for $$\text{Li}^+$$ breaks the equality, so the set is not isoelectronic.

Option B: $$\text{N}^{3-} (10),\; \text{Li}^+ (2),\; \text{Mg}^{2+} (10),\; \text{O}^{2-} (10)$$ — again the $$2$$ electrons of $$\text{Li}^+$$ cause a mismatch.

Option C: $$\text{N}^{3-} (10),\; \text{O}^{2-} (10),\; \text{F}^- (10),\; \text{Na}^+ (10)$$ — every ion here has exactly $$10$$ electrons, so all four are indeed isoelectronic.

Option D: $$\text{Li}^+ (2),\; \text{Na}^+ (10),\; \text{O}^{2-} (10),\; \text{F}^- (10)$$ — the presence of $$\text{Li}^+$$ with only $$2$$ electrons destroys the uniformity.

Clearly, only Option C satisfies the requirement that each species in the list contain the same number of electrons.

Hence, the correct answer is Option C.