The exact volume of $1MNaOH$ solution required to neutralize $50mL of1M{H}_{3}P{O}_3$ solution and $100mL of 2M{H}_{3}P{O}_2$  respectively are

$2NaOH+H_{3}P{O}_3\to N{a}_{2}HP{O}_3+2H_{2}O$

$\begin{array}{l}{M}_b=1M{M}_a=1M\\ V_b=x{V}_a=50mL\\ n_b=2n_a=1\end{array} \frac{M_a{V}_a}{n_a}=\frac{M_b{V}_b}{n_b}\frac{1\times 50}{1}=\frac{1\times x}{2}$

x = 100 mL

$NaOH+H_{3}P{O}_2\to Na{H}_{2}P{O}_2+H_{2}O$

$\begin{array}{l}{M}_b=1M{M}_a=2M\\ V_b=x{V}_a=100\\ n_b=1n_a=1\end{array} \frac{{\mu }_a{V}_a}{n_a}=\frac{{\mu }_b{V}_b}{n_b}\frac{2\times 100}{1}=\frac{1\times x}{1}$

$x=V_b=200 mL$