Answer
$4.05\times10^{3}$ years.
Work Step by Step
Original activity $A_{0}=18.3\,dps$
Present activity $A=11.2\,dps$
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5715\,y}=1.2126\times10^{-4}\,y^{-1}$
$\ln(\frac{A_{0}}{A})=kt$ where $t$ is the age.
$\implies \ln(\frac{18.3}{11.2})=0.491=1.2126\times10^{-4}\,y^{-1}(t)$
Then, $t=\frac{0.491}{1.2126\times10^{-4}\,y^{-1}}=4.05\times10^{3}\,y$
