Answer
2 mg
Work Step by Step
Amount of radionuclide at the beginning $A_{0}=16\,mg$
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{30\,y}=0.0231\,y^{-1}$
Time $t=90\,y$
$\ln(\frac{A_{0}}{A})=kt$ where $A$ is the amount of radionuclide remaining.
$\implies \ln(\frac{16\,mg}{A})=0.0231\times90=2.079$
Taking the inverse $\ln$ of both sides, we have
$\frac{16\,mg}{A}=e^{2.079}=7.996$
Or $A=\frac{16\,mg}{7.996}=2\,mg$
