To neutralize completely $20\text{mL}$ of $0.1\text{M}$ aqueous solution of phosphorous acid (${\text{H}}_3{\text{PO}}_3$), the volume of $0.1\text{M}$ aqueous $\text{KOH}$ solution required is:

Phosphorous acid (${\text{H}}_3{\text{PO}}_3$) is a dibasic acid (contains only two ionizable $P-OH$ bonds). Its $n$ -factor is $2$.
$\text{KOH}$ is a monobasic base, its $n$ -factor is $1$.
Equivalents of acid $=$ Equivalents of base.
$N_1{V}_1(\text{acid})=N_2{V}_2(\text{base})$
$(M\times n{)}_1{V}_1=(M\times n{)}_2{V}_2$
$0.1\times 2\times 20=0.1\times 1\times V_2$.
$4=0.1\times V_2$
$V_2=40\text{mL}$.