Answer
53 days.
Work Step by Step
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{8.0\,d}=0.086625\,d^{-1}$
Original amount $A_{0}=10.\,\mu g$
Amount remaining $A=10.\,\mu g\times\frac{1}{100}=0.10\,\mu g$
Recall that $\ln(\frac{A_{0}}{A})=kt$ where $t$ is the time required.
$\implies \ln(\frac{10.\,\mu g}{0.10\,\mu g})=4.605=0.086625\,d^{-1}(t)$
$\implies t=\frac{4.605}{0.086625\,d^{-1}}=53\,d$
