放射性碳测年（或碳-14测年）是一种能够准确测定大约 60,000 年以内的有机材料年龄的科学方法。该技术基于碳-14同位素的衰变。原始原子数量衰变一半所需的时间定义为半衰期。该过程由方程 $N = N_0 e^{-kt}$ 建模，其中 $k$ 是常数，$t$ 是以年为单位的时间，$N_0$ 是物体中最初的物质含量。 碳-14的半衰期为 5730 年，这意味着在生物体死亡 5730 年后，其一半的碳-14原子已经衰变。

1. Substitute the half-life conditions into the given model

When $t = 5730$, $N = \frac{1}{2}N_0$.
$\frac{1}{2}N_0 = N_0 e^{-5730k}$

2. Solve for $k$ using natural logarithms

$0.5 = e^{-5730k}$
$\ln 0.5 = -5730k$
$k = \frac{\ln 0.5}{-5730}$
$k \approx 0.000120968 \approx 0.000121 \text{ (to 3 s.f.)}$

1. Substitute the given percentage and the value of $k$ from part (a) into the model

From part (a), $k = \frac{\ln 2}{5730}$.
Given $N = 0.08N_0$,
$0.08N_0 = N_0 e^{-\left(\frac{\ln 2}{5730}\right)t}$

2. Solve for $t$ using logarithms

$0.08 = e^{-\left(\frac{\ln 2}{5730}\right)t}$
$\ln 0.08 = -\left(\frac{\ln 2}{5730}\right)t$
$t = -\frac{5730 \ln 0.08}{\ln 2}$
$t \approx 20879.43$

3. State the final answer rounded to the nearest whole number

$t = 20879 \text{ (to nearest whole number)}$