Can you please elaborate on that part? I know that the E value of
Mno4- is +1.51, and the one for BrO3- is +1.48. What does that
suggest?? that Mno4- will more likely to be reduced by Fe+ than
Bro3-?? please explain
The first and foremost hint, even before you calculate the electrode potentials is the solubility rule taught in general chemistry. Hydroxides of most metals are insoluble. Write the equations : Fe(II) + hydroxide ions = iron (II) hydroxide. This is why alkaline permanganate is not a good choice. However, this is not the final answer because you would like to further confirm whether C or D is the correct choice.
Next determine the E_cell= E(cathode)-E(anode) > 0 V or not.
Now you need to oxidize Fe(II) to Fe(III), so the iron half cell will be the anode. As a mnemonic, anode and oxidation both start with vowels; cathode and reduction don't. You would like to reduce the permanganate in an alkaline medium, this half cell is the cathode.
Repeat the same for bromate half cell in acidic medium and Fe(II). Both (C) and (D) would turn out to be positive E_cells. However, the (C) is not an answer because you don't want iron to precipitate out as iron(II) hydroxide.
In real life analysis, Fe(II) can be routinely titrated with potassium permanganate but in highly acidic medium. For practice, check the E_cell after writing a balanced reaction.