Answer
$2.4\times10^{4}\,yr$
Work Step by Step
$R_{0}=15.3\,dis/min\cdot g\,C$
$R=0.85\,dis/min\cdot g\,C$
$k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,yr}=1.21\times10^{-4}/yr$
Recall that $\ln(\frac{R}{R_{0}})=-kt$
$\implies t=-\frac{\ln(\frac{R}{R_{0}})}{k}=-\frac{\ln(\frac{0.85}{15.3})}{1.21\times10^{-4}/yr}=2.4\times10^{4}\,yr$