NO3- + I- ------------> IO3- + NO2

Can someone help me balance this redox reaction??

To balance NO3- + I- ---> IO3- + NO2, split it into half reactions

The balanced half reactions are:
NO3- + 2H+ + e- ---> NO2 + H2O
I- + 3H2O ---> IO3- + 6H+ + 6e-

Now, multiply the first half reaction by 6 to get the 6 electrons needed to use up the 6 electrons produced by the 2nd half reaction:
6NO3- + 12H+ 6e- ---> 6NO2 + 6H2O and add it to the 2nd half reaction
I- + 3H2O ---> IO3- + 6H+ + 6e-
--------------------------------------…
6NO3- + I- + 6H+ ---> IO3- + 6NO2 + 3H20

6NO₃⁻ + I⁻ + 6H+ ------> IO₃⁻ + 6NO₂ + 3H₂O

Just elaborate more about where the oxidation charges:
N + 3(-2) = -1 --> N = 5 e⁻
5 - 4 = 1e⁻
Similarly for I:
-1 - 5 = - 6e⁻