#### Answer

0.178

#### Work Step by Step

Half-life $t_{1/2}$=28.1 years
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{28.1\,y}=0.02466\,y^{-1}$
$t=(2015-1945)\,y=70\,y$
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 $t$.
$\implies \ln(\frac{A}{A_{0}})=-(0.02466\,y^{-1})(70\,y)=-1.7262$
Taking the inverse $\ln$ of both the sides, we have
$\frac{A}{A_{0}}=e^{-1.7262}=0.178$