#### Answer

0.10

#### Work Step by Step

Half-life $t_{1/2}$=30.17 years
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{30.17\,y}=0.02297\,y^{-1}$
$t=100\,y$
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 the sample after time $t$.
$\implies \ln(\frac{A}{A_{0}})=-(0.02297\,y^{-1})(100\,y)=-2.297$
Taking the inverse $\ln$ of both the sides, we have
$\frac{A}{A_{0}}=e^{-2.297}=0.10$