The correct option is A IO−3
Oxidaiton of I− with KMnO4 in alkaline solution is given by
Reaction: 6MnO−4+I−+6OH−→IO−3+6MnO2−4+3H2O
Here the oxidation state of Mn reduced from +7 to +6 and the oxidation state of I oxidised from -1 to +5.
6e− is involved in the reaction.
Oxidation product form is IO−3.
Hence, option A is correct.