1. Identify the basicity of $H_3PO_3$

   Phosphorous acid has structure $HPO(OH)_2$.

   • It contains two ionisable hydrogens (the two $-OH$ hydrogens).
   • The $P-H$ hydrogen is not acidic.

   Therefore, $H_3PO_3$ is a dibasic acid.

   Hence, complete neutralisation is: $H_3PO_3 + 2KOH \rightarrow K_2HPO_3 + 2H_2O$

2. Calculate moles of $H_3PO_3$

   Given:

   • Volume $= 20\,\text{mL} = 0.020\,\text{L}$
   • Molarity $= 0.1\,\text{M}$

   Moles of acid: $n = M \times V = 0.1 \times 0.020 = 0.002\,\text{mol}$

3. Find moles of $KOH$ required

   Since $1$ mole of $H_3PO_3$ requires $2$ moles of $KOH$: $n(\text{KOH}) = 2 \times 0.002 = 0.004\,\text{mol}$

4. Calculate volume of $0.1\,\text{M}$ KOH

   Using $V = \frac{n}{M} = \frac{0.004}{0.1} = 0.040\,\text{L}$

   Convert to mL: $0.040\,\text{L} = 40\,\text{mL}$

5. Check options

   • A: $40\,\text{mL}$ ✅
   • B: $20\,\text{mL}$ ❌
   • C: $10\,\text{mL}$ ❌
   • D: $60\,\text{mL}$ ❌

Therefore, the correct answer is A: $40\,\text{mL}$.