Correct Answer - 3
`NaBrhArr Na^(+)+Br^(-)`
`{:(2H_(2)O,+,2e^(-),rarr,H_(2),+,2OH^(-)),(nA^(+),+,OH^(-),rarr,NaOH,,):}]`at cathode
`Br^(-)+e^(-)rarrBr`
`Br+Brrarr Br_(2)" "(` at anode `)`
So, the products are `H_(2),Br_(2)` and `NaOH,H_(2)` at cathode , `Br_(2)` at anode and `NaOH` in solution.