Answer
0.757 or 75.7%
Work Step by Step
Half-life $t_{1/2}$=5730 years
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,y}=1.2094\times10^{-4}\,y^{-1}$
$t=2300\,y$
Recall that $\ln(\frac{A}{A_{0}})=-k t$ where $A_{0}$ is the amount of sample at the beginning and $A$ is the amount of sample after time $t$.
$\implies \ln(\frac{A}{A_{0}})=-(1.2094\times10^{-4}\,y^{-1})(2300\,y)=-0.278$
Taking the inverse $\ln$ of both the sides, we have
$\frac{A}{A_{0}}=e^{-0.278}=0.757$
