![K_{sp} =\text{[Ca$^{2+}$][OH$^{-}$]}^{2} = x\times (2x)^{2} = 4x^{3} = 6.5 \times 10^{-6}\\x^{3} = \dfrac{6.5\times 10^{-6}} {4} =1.62 \times 10^{-6}\\\\x = \sqrt[3]{1.62 \times 10^{-6}} = \textbf{0.025 mol/L}\\\text{The concentration of Ca$^{2+}$ ions is $\large \boxed{\textbf{ 0.025 mol/L}}$}](/tpl/images/0388/6219/c09ce.png)
Someone me with 84 and explain got to do it : ((
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