Chapter 32 - Nuclear Physics and Nuclear Radiation - Problems and Conceptual Exercises - Page 1155: 81

$2220\ yr$

We know that $R_{1g}=\frac{R}{m}=\frac{1.38Bq}{7.82g}=0.176Bq/g$ Now we can find the required age $t=\frac{1}{\lambda}\ln \frac{R_{0,1g}}{R_{1g}}=\frac{1}{1.21\times 10^{-4}y^{-1}}\ln(\frac{0.231Bq}{0.1765Bq})=2220\ yr$