Answer
$2.9\times10^{4}$ years ago.
Work Step by Step
Recall: $\ln\frac{A_{t}}{A_{0}}=-\frac{0.693}{t_{1/2}}\times t$
$\implies \ln \frac{0.48}{15.3}=-\frac{0.693}{5715\,y}\times t$
$\implies -3.461822=-1.212598\times10^{-4}\,y^{-1}\times t$
$\implies t= \frac{-3.461822}{-1.212598\times10^{-4}\,y^{-1}}$
$=2.9\times10^{4}\,y$