1. Write the dissolution equilibrium

Calcium phosphate is: $\mathrm{Ca_3(PO_4)_2}$

Its dissolution in water is: $\mathrm{Ca_3(PO_4)_2(s) \rightleftharpoons 3Ca^{2+} + 2PO_4^{3-}}$

So, if its molar solubility is $s$ mol L$^{-1}$, then: $[\mathrm{Ca^{2+}}]=3s, \qquad [\mathrm{PO_4^{3-}}]=2s$

2. Write the solubility product expression

$K_{sp}=[\mathrm{Ca^{2+}}]^3[\mathrm{PO_4^{3-}}]^2$

Substitute the ion concentrations: $K_{sp}=(3s)^3(2s)^2=27s^3 \cdot 4s^2=108s^5$

Thus, $K_{sp}=108s^5$

3. Convert given solubility into molar solubility

Given solubility = $W$ g per $100$ mL.

Therefore, in $1$ L (= $1000$ mL), solubility is: $10W \text{ g L}^{-1}$

Hence molar solubility: $s=\frac{10W}{M}$

4. Substitute into $K_{sp}$

$K_{sp}=108\left(\frac{10W}{M}\right)^5$

Now, $108\times 10^5 \left(\frac{W}{M}\right)^5 = 1.08\times 10^7 \left(\frac{W}{M}\right)^5$

Approximately, $K_{sp} \approx 10^7\left(\frac{W}{M}\right)^5$

5. Match with the options

This corresponds to:

Option C: $10^7\left(\frac{W}{M}\right)^5$