Solubility of calcium phosphate (molecular mass, $\mathrm{M}$) in water is ${\mathrm{W}}_{\mathrm{g}}$ per $100 \mathrm{mL}$ at $25°\mathrm{C}.$ Its solubility product at $25°\mathrm{C}$ will be approximately.

Given the solubility of ${\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2$ is $\mathrm{Wg}/100 \mathrm{mL}$. The solubility in $\mathrm{mol}/\mathrm{L}$ is $\mathrm{S}=\frac{\mathrm{W}}{\mathrm{M}}\times \frac{1000}{100}\mathrm{mol}/\mathrm{L}$.

The equilibrium reaction of calcium phosphate is 

$\begin{array}{cccccc}& {\mathrm{Ca}}_3{\left({\mathrm{PO}}_4\right)}_2& \stackrel{}{\rightleftharpoons }& 3{\mathrm{Ca}}^{2+}& +& 2{\mathrm{PO}}_4^{3-}\\ \mathrm{At} \mathrm{equilibrium}& & & 3\mathrm{S}& & 2\mathrm{S}\end{array}$

Now, ${\mathrm{K}}_{\mathrm{sp}}={\left[{\mathrm{Ca}}^{2+}\right]}^3{\left[{\mathrm{PO}}_4^{3-}\right]}^2$

$\Rightarrow {\mathrm{K}}_{\mathrm{sp}}={\left(3\mathrm{S}\right)}^3{\left(2\mathrm{S}\right)}^2=108\times {\mathrm{S}}^5$

$$\begin{gathered} \Rightarrow {\mathrm{K}}_{\mathrm{sp}}=108\times {\left(\frac{\mathrm{W}}{\mathrm{M}}\times 10\right)}^5 \\ \Rightarrow {\mathrm{K}}_{\mathrm{sp}}=108\times {10}^5\times {\left(\frac{\mathrm{W}}{\mathrm{M}}\right)}^5 \end{gathered}$$

$=1.08\times {10}^7{\left(\frac{\mathrm{W}}{\mathrm{M}}\right)}^5$