For a first-order reaction, the number of half-lives is given by $n = \frac{t}{t_{1/2}}$. Given total time $t = 8000 \text{ years}$ and half-life $t_{1/2} = 2000 \text{ years}$, the number of half-lives $n = \frac{8000}{2000} = 4$. The concentration of the reactant remaining after $n$ half-lives is given by the formula: $[A] = \frac{[A]_0}{2^n}$ Where: $[A] = \text{final concentration} = 0.02 \text{ M}$ $[A]_0 = \text{initial concentration}$ Substituting the values into the formula: $0.02 = \frac{[A]_0}{2^4}$ $0.02 = \frac{[A]_0}{16}$ $[A]_0 = 0.02 \times 16 = 0.32 \text{ M}$. Therefore, the initial concentration was $0.32 \text{ M}$.