The molar solubility of $\mathrm{Zn}(\mathrm{OH})_2$ in 0.1 M NaOH solution is $x \times 10^{-18} \mathrm{M}$. The value of $x$ is $\_\_\_\_$ (Nearest integer)
(Given : The solubility product of $\mathrm{Zn}(\mathrm{OH})_2$ is $2 \times 10^{-20}$ )
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