NO3 + 12 + 103 + NO2

What is being oxidized?

perhaps is NO3

this question is somehow not correct

NO₃⁻ is reduced; I₂ is oxidized

Explanation:

I think your equation is supposed to be

NO₃⁻ + I₂ ⟶ IO₃⁻ + NO₂

If you break the equation into two half-reactions, you get

NO₃⁻ + 2H⁺ + e⁻ ⟶ NO₂ + H₂O

I₂ + 3H₂O ⟶ IO₃⁻ + 6H⁺ + 5e⁻

We see that NO₃⁻ gains an electron, so it is reduced.

I₂ loses five electrons, so it is oxidized.