Answer
$1.95\times10^{4} \,y$
Work Step by Step
Amount of radionuclide at the beginning $A_{0}=8.50\,\mu g$
Amount of radionuclide remaining
$A=0.80\,\mu g$
Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,y}=1.21\times10^{-4}\,y^{-1}$
$\ln(\frac{A_{0}}{A})=kt$ where $t$ is the age.
$\implies \ln(\frac{8.50}{0.80})=2.36321=(1.21\times10^{-4}\,y^{-1})\times t$
Or $t= \frac{2.36321}{1.21\times10^{-4}y^{-1}}=1.95\times10^{4} \,y$
