Explanation

(c)
$H_{3}{PO}_{3}$ is a dibasic acid (containing two ionisable protons attached to O directly).
$\ \ \ \ \ \ \ \ \ \ \ \ \ H_{3}{PO}_{3}\ \rightleftharpoons \ 2H^{+} + \ HP{O_{3}}^{2 -}$
$\therefore\ \ \ 0.1\ M\ \ H_{3}PO_{3} = \ 0.2\ NH_{3}\ PO_{3}$
and 0.1 M KOH = 0.1N KOH
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ N_{1}V_{1} = \ N_{2}V_{2}$
(KOH) ($H_{3}PO_{3})$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.1V_{1}\ = \ 0.2\ \times \ 20\$
${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ V}_{1}$ = 40mL