1. Write the dissolution equilibrium

$Zn(OH)_2(s) \rightleftharpoons Zn^{2+} + 2OH^-$

Given,

$K_{sp} = [Zn^{2+}][OH^-]^2 = 2 \times 10^{-20}$

2. Effect of common ion from $0.1\,M\ NaOH$

Since $NaOH$ is a strong base,

$[OH^-] \approx 0.1\,M$

Let the molar solubility of $Zn(OH)_2$ be $s$. Then,

$[Zn^{2+}] = s$

Because $0.1\,M$ is much larger than the contribution $2s$ from dissolution, we take

$[OH^-] \approx 0.1$

3. Apply solubility product expression

$K_{sp} = s(0.1)^2$

So,

$s = \frac{2 \times 10^{-20}}{(0.1)^2}$

$s = \frac{2 \times 10^{-20}}{10^{-2}} = 2 \times 10^{-18}\,M$

4. Match with the given form

The solubility is given as

$x \times 10^{-18}\,M$

Comparing,

$x = 2$

5. Comparison with stored answer

Stored correct answer = 2, which matches our result.