The volume of $0.1 M$ oxalic acid that can be completely oxidized by $20 ml$ of $0.025 M KMn{O}_4$ solution is-

$$\begin{gathered} \underset{\mathrm{RA}}{{\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{O}}_4\to {\mathrm{CO}}_2} ; \underset{\mathrm{OA}}{ {\mathrm{KMnO}}_4\to {\mathrm{Mn}}^{+2}} \\ {n}_f = 2 n_f = 5 \\ {m}_{eq}{H}_2{C}_2{O}_4 = m_{eq}KMn{O}_4 \\ M \times V \times n_f = M \times V \times n_f \\ 0.1 \times V \times 2 = 0.025 \times 20 \times 5 \\ V = 12.5 ml \end{gathered}$$