Answer
91.7%
Work Step by Step
Half-life of $\,\,^{131}I$=8.02 days
Decay constant $\lambda=\frac{0.693}{t_{1/2}}=\frac{0.693}{8.02\,d}=0.086409\,d^{-1}$
$t=1\,d$
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 1 day.
$\implies \ln(\frac{A}{A_{0}})=-(0.086409\,d^{-1})(1\,d)=-0.086409$
Taking the inverse $\ln$ of both the sides, we have
$\frac{A}{A_{0}}=e^{-0.086409}=0.917$
Percentage of the sample remaining=$0.917\times100\%=91.7\%$
