Calculate the molar solubility of fe(oh)2 in pure water. (the value of ksp for fe(oh)2 is 4.87×10−17.)

When we have this balanced equation for a reaction:
Fe(OH)2(s) ↔ Fe+2  + 2OH-

when Fe(OH)2 give 1 mole of Fe+2 & 2 mol of OH-
so we can assume [Fe+2] = X and [OH-] = 2 X
 when Ksp = [Fe+2][OH-]^2
and have Ksp = 4.87x10^-17 
[Fe+2]= X
[OH-] = 2X
so by substitution 
4.87x10^-17 = X*(2X)^2
∴X^3 = 4.8x10^-17 / 4
∴the molar solubility X = 2.3x10^-6 M

The answer is true for your question.