The solubility of Ag3PO4 is 2.1x10-4 g/100g water.?

1)
Ag₃PO₄(s) ⇌ 3Ag⁺(aq) + PO₄³⁻(aq)

2)
Density of water = 1 g/mL
Volume of the solution = (100 g) × (1 g/mL) = 100 mL = 0.1 L
Molar mass of Ag₃PO₄ = (107.9×3 + 31.0 + 16.0×4) g/mol = 418.7 g/mol
Molarity = [(2.1 × 10⁻⁴)/418.7 mol] / (0.1 L) = 5.0 × 10⁻⁶ M

3)
Ksp = [Ag⁺]³ [PO₄³⁻] = (3 × 5.0 × 10⁻⁶)³ × (5.0 × 10⁻⁶) = 1.7 × 10⁻²⁰