At 25° C, Kb for BOH $=1.0\times {10}^{-12}\cdot 0.01\mathrm{M}$ solution of BOH has $\left[{\mathrm{OH}}^{-}\right]$] :

The reaction takes place as,

$\mathrm{BOH}\rightleftharpoons {\mathrm{B}}^{+}+{\mathrm{OH}}^{-}$

The given values are,

$K_b=1.0\times {10}^{-12}$

$\left[\mathrm{BOH}\right]=0.01\mathrm{M}$

For the equilibrium condition,

${\mathrm{K}}_{\mathrm{b}}=\frac{{\left[{\mathrm{OH}}^{-}\right]}^2}{\left[\mathrm{BOH}\right]}$

Thus,  $1.0\times {10}^{-12}=\frac{{\left[{\mathrm{OH}}^{-}\right]}^2}{0.01}$

$\begin{array}{r}{\left[{\mathrm{OH}}^{-}\right]}^2=1\times {10}^{-14}\\ \left[{\mathrm{OH}}^{-}\right]=1.0\times {10}^{-7}\mathrm{mol} \mathrm{L}\end{array}$