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The expression for $K_{sp}$ for this reaction is:
$K_{sp}= [Mn^{2+} ][OH^-]^2$.
Substituting the given values:
$1.6 times 10^{-13} = [Mn^{2+}] (0.020)^2$
Solving for $[Mn^{2+}]$:
$[Mn^{2+}] = frac{1.6 times 10^{-13}}{(0.020)^2} = boxed{4.0 times 10^{-10} M}$