Answer
8650 years.
Work Step by Step
2690 decays/min is 44.83 decays/second.
The activity of the sample is $\frac{44.83Bq}{0.500kg}=89.7\frac{Bq}{kg}$.
In atmospheric carbon, the corresponding figure is 255 Bq/kg, as stated in Example 43.9.
The age of the sample is $t=-\frac{ln(89.7/255)}{\lambda}$, where $\lambda$ is given for carbon in Example 43.9.
$t=2.73\times10^{12}s\approx8650yr$
