To convert 12 g of NaH₂PO₄ completely into Na₃PO₄, the volume of 1molar NaOH required is

${\mathrm{NaH}}_2{\mathrm{PO}}_4\left(\mathrm{s}\right)+2\mathrm{NaOH}\left(\mathrm{aq}\right)\to {\mathrm{Na}}_3{\mathrm{PO}}_4\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

no. of moles of NaH₂PO₄=$\frac{12}{120}=0.1\mathrm{moles}$

0.1 moles of NaH₂​PO₄​ requires 0.2 moles of NaOH.

Given,

1 mole NaOH is present in 1000 ml of solution.

Therefore, 0.2 moles of NaOH ⟶ 0.2×1000.

 Hence, the volume required = 200 ml.