Answer
14.0 %
Work Step by Step
Half-life of $\,\,^{14}C$=5715 years.
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{5715\,y}=1.2126\times10^{-4}\,y^{-1}$
$t=16230\,y$
Recall that $\ln(\frac{N}{N_{0}})=-kt$ where $N_{0}$ is the amount of carbon-14 at the beginning and $N$ is the amount after 16230 years.
$\implies \ln(\frac{N}{N_{0}})=-(1.2126\times10^{-4}\,y^{-1})(16230\,y)=-1.968$
Taking the inverse $\ln$ of both the sides, we have
$\frac{N}{N_{0}}=e^{-1.968}=0.140$
Percentage of the sample remaining=$\frac{N}{N_{0}}\times100\%=0.140\times100\%=14.0\%$
