The volume of 0.025M ${\displaystyle Ca{\left(OH\right)}_2}$ solution which can neutralise 100 ml of ${\displaystyle {10}^{-4}M{H}_{3}P{O}_4}$ is

$2H_{3}P{O}_4+3Ca{\left(OH\right)}_2\to C{a}_3{\left(P{O}_4\right)}_2+6H_{2}O$

$$\begin{gathered} {\mathrm{V}}_1{\mathrm{N}}_1={\mathrm{V}}_2{\mathrm{N}}_2 \\ {\mathrm{V}}_2 = \frac{{\mathrm{N}}_1{\mathrm{V}}_1}{{\mathrm{N}}_2}=\frac{3\times {10}^{-4}\times 100 \mathrm{ml}}{0.05}=0.6 \mathrm{ml} \end{gathered}$$

Hence the correct option is (C). 0.6 ml