The species in which the N atom is in a state of sp hybridization is:

We begin by recalling the rule that decides hybridisation. The steric number is given by

$$\text{Steric number}= (\text{Number of } \sigma\text{-bonds on the central atom}) + (\text{Number of lone pairs on the central atom}).$$

If the steric number is $$2$$, the hybridisation is $$sp$$; if the steric number is $$3$$, the hybridisation is $$sp^2$$; if the steric number is $$4$$, the hybridisation is $$sp^3$$, and so on.

Now we examine each species one by one, always keeping nitrogen as the central atom.

1. The ion $$NO_3^-$$ (nitrate)

Total valence electrons:

$$N = 5,\; 3O = 3 \times 6 = 18,\; \text{extra charge} = 1.$$

So

$$5 + 18 + 1 = 24\text{ electrons} = 12\text{ pairs}.$$

The best Lewis structure gives nitrogen bonded to three oxygens through three $$\sigma$$-bonds. Nitrogen has no lone pair in this structure.

Therefore

$$\text{Steric number} = 3\sigma + 0\text{ lone} = 3 \implies \text{hybridisation } sp^2.$$

Hence nitrogen is not $$sp$$ here.

2. The molecule $$NO_2$$ (nitrogen dioxide)

Total valence electrons:

$$N = 5,\; 2O = 12,\; \text{charge} = 0.$$

Thus

$$5 + 12 = 17\text{ electrons}.$$

The odd number means one electron remains unpaired. Nitrogen is attached to two oxygens through two $$\sigma$$-bonds, and one lone electron (not a pair) still counts as one region of electron density, equivalent to one lone pair for geometry purposes.

So

$$\text{Steric number} = 2\sigma + 1\text{ lone region} = 3 \implies sp^2.$$

Nitrogen is again not $$sp$$.

3. The ion $$NO_2^+$$ (nitronium)

Total valence electrons:

$$N = 5,\; 2O = 12,\; \text{positive charge} = -1.$$

Thus

$$5 + 12 - 1 = 16\text{ electrons} = 8\text{ pairs}.$$

The optimum Lewis structure has nitrogen doubly bonded to each oxygen:

$$O=N=O$$

This gives nitrogen exactly two $$\sigma$$-bonds and no lone pairs.

Therefore

$$\text{Steric number} = 2\sigma + 0\text{ lone} = 2 \implies \boxed{sp\text{ hybridisation}}.$$

So in $$NO_2^+$$ the nitrogen atom is $$sp$$ hybridised.

4. The ion $$NO_2^-$$ (nitrite)

Total valence electrons:

$$N = 5,\; 2O = 12,\; \text{negative charge} = +1.$$

Thus

$$5 + 12 + 1 = 18\text{ electrons} = 9\text{ pairs}.$$

The preferred Lewis structure shows nitrogen bonded to two oxygens through two $$\sigma$$-bonds and possessing one lone pair.

Hence

$$\text{Steric number} = 2\sigma + 1\text{ lone} = 3 \implies sp^2.$$

Nitrogen is not $$sp$$ here either.

Comparing all four species, only $$NO_2^+$$ supplies a steric number of $$2$$ and therefore places nitrogen in the $$sp$$ state.

Hence, the correct answer is Option C.