Answer
75.8 days
Work Step by Step
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{11.4\,d}=0.06078947\,d^{-1}$
Let the current activity be $A_{0}=100$
Then, activity after time $t$, $A=1\%\,of\,A_{0}=1$
Recall that $\ln(\frac{A_{0}}{A})=kt$ where $t$ is the time required.
$\implies \ln(\frac{100}{1})=4.60517=0.06078947\,d^{-1}(t)$
Or $ t=\frac{4.60517}{0.06078947\,d^{-1}}=75.8\,d$
