${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}+{\text{6H}}_{\text{2}}\text{O}\to {\text{4H}}_{\text{3}}{\text{PO}}_{\text{3}}$ ......(i)

Neutralisation :

${\text{H}}_{\text{3}}{\text{PO}}_{\text{3}}\text{+}\text{2NaOH}\to \left[{\text{Na}}_{\text{2}}{\text{HPO}}_{\text{3}}\text{+}{\text{2H}}_{\text{2}}\text{O}\right]\text{×}\text{4}$ .......(ii)

Adding Eqs. (i) and (ii)

$\begin{array}{ccccc}{\text{P}}_4{\text{O}}_6& +& 8\text{NaOH}& ⟶& 4{\text{Na}}_2{\text{HPO}}_3+2{\text{H}}_2\text{O}\\ \mathrm{1 }\text{mol}& & \mathrm{8 }\text{mol}& & \end{array}$ .....(iii)

Number of moles of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}\text{,}$

$\text{n}=\frac{\text{m}}{\text{M}}=\frac{1\text{.}1}{220}=\frac{1}{200}\text{ mol}$

(Molar mass ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}=(4\times 31)+(6\times 16)=220$

$\because$   Product formed by $\text{1}$ mole of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ is neutralised by $\text{8}$ moles $\text{NaOH}$

$\therefore$   Product formed by $\frac{1}{200}$ moles of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ will be neutralised by $\text{NaOH}$

                    $=8\times \frac{1}{200}=\frac{8}{200}\text{ moles NaOH}$

Given, Molarity of $\text{NaOH}\text{=}\text{0}\text{.1}\text{M}\text{=}\text{0}\text{.1}\text{mol/L}$

                     $\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume in litres}}$

                     $\text{Volume}=\frac{\text{Number of moles}}{\text{Molarity}}$

                                 $=\frac{8}{200}\times \frac{1}{0\text{.}1}=0\text{.}4\text{ L}\text{ or }400\text{ mL}$

∴   $\text{400}\text{mL}\text{NaOH}$ is required.