Explanation

(d)
$KMnO_{4}$ reacts with oxalic acid according to the following equation.
$2MnO_{4}^{-} + 5C_{2}O_{4}^{2 -} + 16H^{+} \rightarrow 2Mn^{2 +} + 10CO_{2} + 8H_{2}O$
Eq. mass of $KMnO_{4} = \frac{mol.mass}{7 - 2}$
$N_{KMnO_{4}} = 5 \times$molarity=$5 \times 10^{- 4}$
Eq. mass of $C_{2}O_{4}^{2 -} = \frac{mol.\ \ mass}{2(4 - 3)} = \frac{mol.\ \ mass}{2}$
$N_{C_{2}O_{4}^{2 -}} = 2 \times$ molarity=$2 \times 10^{- 2}$
$N_{1}V_{1} = N_{2}V_{2}$
$5 \times 10^{- 4} \times V_{1} = 2 \times 10^{- 2} \times 0.5$
$V_{1} = \frac{2 \times 10^{- 2} \times 0.5}{5 \times 10^{- 4}} = 20\ L$