1. Write the dissolution equilibrium

For lead sulfide: $\text{PbS}(s) \rightleftharpoons \text{Pb}^{2+}(aq) + \text{S}^{2-}(aq)$

If its solubility in pure water is $s$ mol L$^{-1}$, then at equilibrium: $[\text{Pb}^{2+}] = s, \qquad [\text{S}^{2-}] = s$

2. Use the solubility product expression

Given: $K_{sp} = [\text{Pb}^{2+}][\text{S}^{2-}] = s \cdot s = s^2$

So, $s^2 = 8 \times 10^{-28}$

$s = \sqrt{8 \times 10^{-28}}$

3. Simplify the square root

$s = \sqrt{8} \times 10^{-14}$

Now, $\sqrt{8} = 2\sqrt{2}$

Given $\sqrt{2} = 1.41$, $\sqrt{8} = 2 \times 1.41 = 2.82$

Hence, $s = 2.82 \times 10^{-14}\ \text{mol L}^{-1}$

4. Match with the required form

We need: $s = x \times 10^{-16}\ \text{mol L}^{-1}$

Since $2.82 \times 10^{-14} = 282 \times 10^{-16}$

therefore, $x = 282$

5. Nearest integer

$\boxed{282}$

6. Comparison with stored answer

Stored correct answer = $282$.

Our derived answer also is $282$, so they agree.