Answer
0.062
Work Step by Step
Half-life $t_{1/2}=7.0\times10^{8}\,y$
Decay constant $\lambda=\frac{0.693}{t_{1/2}}=\frac{0.693}{7.0\times10^{8}\,y}=9.9\times10^{-10}\,y^{-1}$
$t=2.8\times10^{9}\,y$
Recall that $\ln(\frac{A}{A_{0}})=-\lambda t$ where $A_{0}$ is the amount of sample at the beginning and $A$ is the amount of sample after after time $t$.
$\implies \ln(\frac{A}{A_{0}})=-(9.9\times10^{-10}\,y^{-1})(2.8\times10^{9}\,y)=-2.772$
Taking the inverse $\ln$ of both the sides, we have
$\frac{A}{A_{0}}=e^{-2.772}=0.062$
