What is the molar solubility of barium fluoride ( baf2 ) in water? the solubility-product constant for baf2 is 1.7 × 10-6 at 25∘c?

Molar solubility is the number of moles that are dissolved in 1 L solution. 
when BaF₂ dissolves it dissociates into the following ions
BaF₂ --> Ba²⁺ + 2F⁻
if the molar solubility of BaF₂ is X, then molar solubility of Ba²⁺ is X and F⁻ is 2x
then the formula for the solubility product constant -ksp is;
ksp = [Ba²⁺][F⁻]²
ksp = X * (2X)²
ksp = 4X³
since ksp = 1.7 x 10⁻⁶
4X³ = 1.7 x 10⁻⁶
X = 0.0075 M
molar solubility of BaF₂ is 0.0075 M