To determine the molar solubility (s) of Ba3(PO4)2 in terms of Ksp, we need to first understand the concept of solubility product constant (Ksp).

1. Solubility Product Constant (Ksp):
Ksp is a constant that represents the equilibrium constant for the dissolution of a sparingly soluble salt in water. It is the product of the concentrations of the ions raised to the power of their stoichiometric coefficients in the balanced chemical equation.

2. Balanced Chemical Equation:
The balanced chemical equation for the dissolution of Ba3(PO4)2 can be written as follows:
Ba3(PO4)2 ⇌ 3Ba2+ + 2PO4^3-

3. Solubility (s):
Solubility (s) refers to the maximum amount of a solute that can dissolve in a given solvent under specific conditions. In this case, we are interested in the molar solubility, which is the number of moles of the solute (Ba3(PO4)2) that can dissolve per liter of solution.

4. Stoichiometry:
Based on the balanced chemical equation, for every mole of Ba3(PO4)2 that dissolves, it produces 3 moles of Ba2+ ions and 2 moles of PO4^3- ions.

Now, let's proceed with the solution:

- The solubility product constant (Ksp) expression for Ba3(PO4)2 can be written as:
Ksp = [Ba2+]^3[PO4^3-]^2

- Since the molar solubility (s) of Ba3(PO4)2 is defined as the concentration of Ba2+ or PO4^3- ions when the compound is in equilibrium with its saturated solution, we can substitute [Ba2+] and [PO4^3-] with 's' in the Ksp expression.

- So, the Ksp expression becomes:
Ksp = s^3 * s^2 = s^5

- Rearranging the equation, we can solve for 's':
s = (Ksp)^(1/5)

Therefore, the molar solubility (s) of Ba3(PO4)2 in terms of Ksp is given by the expression s = (Ksp)^(1/5).

Option D, s = [Ksp/108]^(1/5), is the correct answer as it represents the molar solubility of Ba3(PO4)2 in terms of Ksp.