Answer
1.0 mg
Work Step by Step
Amount of radionuclide at the beginning $A_{0}=16\,mg$
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{4.5\,d}=0.154\,d^{-1}$
Time $t=18\,d$
$\ln(\frac{A_{0}}{A})=kt$ where $A$ is the amount of radionuclide remaining.
$\implies \ln(\frac{16\,mg}{A})=0.154\times18=2.772$
Taking the inverse $\ln$ of both sides, we have
$\frac{16\,mg}{A}=e^{2.772}=16.0$
Or $A= \frac{16\,mg}{16.0}=1.0\,mg$