Answer
18.7 %
Work Step by Step
Half-life $t_{1/2}=28.9\,y$
Decay constant $\lambda=\frac{0.693}{t_{1/2}}=\frac{0.693}{28.9\,y}=0.02398\,y^{-1}$
$t=(2015-1945)y= 70\,y$
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 sample after time $t$.
$\implies \ln(\frac{A}{A_{0}})=-(0.02398\,y^{-1})(70\,y)=-1.6786$
Taking the inverse $\ln$ of both the sides, we have
$\frac{A}{A_{0}}=e^{-1.6786}=0.1866$
Percentage of the sample remaining=$0.1866\times100\%=18.7\%$
