If the solubility of $C{a}_3(P{O}_4{)}_2$ in water is ‘X’ $mo{l}^{–1}$, its solubility product in $mo{l}^5{L}^{–5}$ is

 $C{a}_3(P{O}_4{)}_2 \rightleftharpoons 3C{a}^{2+}+2P{O}_4^{3-}$

$\left[C{a}^{2+}\right]$ = 3X, $\left[P{O}_4^{3-}\right]$ =

$K_{s}p$= [$C{a}^{2+}{]}^3{[PO{]}^4}^{3-}]$

= $\left(3X{)}^3\right(2X{)}^2$

= $\left(27X^3\right)\left(4X^2\right)$

= $108X^5$.

Hence, the correct option is option (D) $108X^5$.