To determine the solubility product (K_sp) of calcium phosphate, we need to consider its chemical formula and how it dissociates in water. The formula for calcium phosphate is $\text{Ca}_3(\text{PO}_4)_2$. When dissolved in water, it dissociates according to the following equation:

$$ \text{Ca}_3(\text{PO}_4)_2 (s) \rightleftharpoons 3 \text{Ca}^{2+} (aq) + 2 \text{PO}_4^{3-} (aq) $$

Let $S$ be the molar solubility of calcium phosphate in $\text{mol}/L$. According to the stoichiometry of the dissociation, if $S$ moles per liter of calcium phosphate dissolve, $3S$ moles per liter of calcium ions and $2S$ moles per liter of phosphate ions are produced.

The expression for the solubility product constant (Ksp) for calcium phosphate in water is the product of the concentrations of the ions raised to the power of their respective coefficients in the balanced equation:

$$ K_{sp} = [\text{Ca}^{2+}]^3 [\text{PO}_4^{3-}]^2 $$

Substituting the concentrations with $3S$ for $\text{Ca}^{2+}$ and $2S$ for $\text{PO}_4^{3-}$, we get:

$$ K_{sp} = (3S)^3 (2S)^2 = 27S^3 \cdot 4S^2 = 108S^5 $$

If the solubility of calcium phosphate is $\text{W}_{g}$ per $100 \text{ mL}$ of water at $25^\circ\text{C}$, we need to convert grams to moles to find $S$. Solubility in moles per liter (which is the same as moles per $1000 \text{ mL}$) is:

$$ S = \frac{\text{W}}{\text{M}} \times \frac{1000 \text{ mL}}{100 \text{ mL}} = 10\frac{\text{W}}{\text{M}} $$

Now, substituting $S$ with $10\frac{\text{W}}{\text{M}}$ into the expression for $K_{sp}$:

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

This simplifies to:

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

Since $108$ is on the order of $10^2$, we can approximate this to:

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

Therefore, the correct answer based on the options provided is:

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