Answer
3.13%
Work Step by Step
Half-life of $\,\,^{14}C$=5730 years.
Decay constant $\lambda=\frac{0.693}{t_{1/2}}=\frac{0.693}{5730\,y}=1.2094\times10^{-4}\,y^{-1}$
$t=28650\,y$
Recall that $\ln(\frac{A}{A_{0}})=-\lambda t$ where $A_{0}$ is the amount of carbon-14 at the beginning and $A$ is the amount after 28650 years.
$\implies \ln(\frac{A}{A_{0}})=-(1.2094\times10^{-4}\,y^{-1})(28650\,y)=-3.465$
Taking the inverse $\ln$ of both the sides, we have
$\frac{A}{A_{0}}=e^{-3.465}=0.0313$
Percentage of the sample remaining=
$\frac{A}{A_{0}}\times100\%=0.0313\times100\%=3.13\%$
