1. Write the dissolution equilibrium

   $Al(OH)_3(s) \rightleftharpoons Al^{3+} + 3OH^-$

2. Write the solubility product expression

   $K_{sp} = [Al^{3+}][OH^-]^3$

   Given: $K_{sp} = 2.4 \times 10^{-24}$

3. Use the common ion effect

   The solution already contains $0.2\,M$ $NaOH$, so $[OH^-] \approx 0.2$

   Let the molar solubility of $Al(OH)_3$ be $s$.

   Then, $[Al^{3+}] = s$

   Since $0.2$ is much larger than $3s$, we take $[OH^-] \approx 0.2$

4. Substitute into the $K_{sp}$ expression

   $2.4 \times 10^{-24} = s(0.2)^3$

   Now, $(0.2)^3 = 0.008 = 8 \times 10^{-3}$

   So, $s = \frac{2.4 \times 10^{-24}}{8 \times 10^{-3}}$

5. Calculate $s$

   $s = \frac{2.4}{8} \times 10^{-24+3}$ $s = 0.3 \times 10^{-21}$ $s = 3 \times 10^{-22} \text{ M}$

6. Match with the options

   $\boxed{3 \times 10^{-22}}$

   This corresponds to Option A.

Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So the answer agrees with the stored correct answer.