${\mathrm{KMnO}}_4\text{ reacts with oxalic acid according to the }$equation,

$2{\mathrm{MnO}}_4^{-}+5{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+16{\mathrm{H}}^{+}⟶$$2{\mathrm{Mn}}^{2+}+10{\mathrm{CO}}_2+8{\mathrm{H}}_2\mathrm{O}$

Here, 20 mLof 0.1 M ${\mathrm{KMnO}}_4$ is equivalent to

$2{\mathrm{MnO}}_4^{-}+5{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+16{\mathrm{H}}^{+}⟶2{\mathrm{Mn}}^{2+}+10{\mathrm{CO}}_2+8{\mathrm{H}}_2\mathrm{O}$

$20 \mathrm{mL}\text{ of }0.1{\mathrm{MKMnO}}_4=20\times 0.1=2\mathrm{m} \mathrm{mol}$

$\because 2\text{ m mol of }{\mathrm{KMnO}}_4\equiv 5\text{ m mol of }{\mathrm{C}}_2{\mathrm{O}}_4^{2-}$

$50 \mathrm{mL}\text{ of }0.1{\mathrm{MH}}_2{\mathrm{C}}_2{\mathrm{O}}_4=50\times 0.1=5\mathrm{m} \mathrm{mol}$

$\text{ Hence, }20 \mathrm{mL}\text{ of }0.1{\mathrm{MKMnO}}_4\equiv 50 \mathrm{mL}\text{ of }0.1{\mathrm{MH}}_2{\mathrm{C}}_2{\mathrm{O}}_4$