Solubility of $B{\left(OH\right)}_2$ in water at ${25}^{0}C$ is ${10}^{-7} M.$ The $K_{sp}$ of $B{\left(OH\right)}_2$ is

$B{\left(OH\right)}_2\left(S\right)\underset{ }{\overset{ }{\rightleftharpoons }}\underset{{10}^{-7}}{B^{2+\left(aq\right)}}+\underset{2\times {10}^{-7}+x}{2O{H}^{-}\left(aq\right)}$

Where $x$ is $\left[O{H}^{-}\right]$ from water.

$H_{2}O\left(l\right)\underset{ }{\overset{ }{\rightleftharpoons }}\underset{x}{H^{+}}+\underset{x+2\times {10}^{-7}}{O{H}^{-}}$

$k_w=\left[H^{+}\right]\left[O{H}^{-}\right]$

${10}^{-14}=x(x+2\times {10}^{-7})\Rightarrow x=\frac{-2\times {10}^{-7}\pm \sqrt{4\times {10}^{-4}+4\times {10}^{-14}}}{2}$

$=\frac{-2\times {10}^{-7}\pm 2\sqrt{2}\times {10}^{-7}}{2}=(\sqrt{2}-1)\times {10}^{-7}M$

$K_{sp}\text{ of }B{\left(OH\right)}_2={10}^{-7}{(2\times {10}^{-7}+(\sqrt{2}-1){10}^{-7})}^2={10}^{-7}{(\sqrt{2}+1)}^2{10}^{-14}=5.827\times {10}^{-21}{M}^3$