Answer
$2.4\times10^{4}\,y$
Work Step by Step
Recall: $\ln\frac{A_{t}}{A_{0}}=-\frac{0.693}{t_{1/2}}\times t$
$\implies \ln \frac{0.85}{15.3}=-\frac{0.693}{5715\,y}\times t$
$\implies -2.89037=-1.212598\times10^{-4}\,y^{-1}\times t$
$\implies t= \frac{-2.89037}{-1.212598\times10^{-4}\,y^{-1}}$
$=2.4\times10^{4}\,y$
