Answer
9.0 days
Work Step by Step
Original amount $A_{0}=4.8\,mg$
Present amount $A=1.2\,mg$
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{4.5\,d}=0.154\,d^{-1}$
$\ln(\frac{A_{0}}{A})=kt$
$\implies \ln(\frac{4.8}{1.2})=1.386=0.154\,d^{-1}(t)$
Then, $t=\frac{1.386}{0.154\,d^{-1}}=9.0\,d$