Answer
23.0 %
Work Step by Step
Decay constant $\lambda=7.95\times10^{-3}\,d^{-1}$
$t=185\,d$
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 $t$.
$\implies \ln(\frac{A}{A_{0}})=-(7.95\times10^{-3}\,d^{-1})(185\,d)=-1.47$
Taking the inverse $\ln$ of both the sides, we have
$\frac{A}{A_{0}}=e^{-1.47}=0.230$
Percentage of the sample remaining=$0.230\times100\%=23.0\%$
