Answer
0.0019 mg
Work Step by Step
The amount of radionuclide at the beginning is $A_{0}=0.56\,mg$.
Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{44.5\,day}=0.01557/day$
Time $t=1 \,year=365\,days$
$\ln(\frac{A_{0}}{A})=kt$ where $A$ is the amount of radionuclide remaining.
$\implies \ln(\frac{0.56\,mg}{A})=0.015573\times365=5.684$
Taking the inverse $\ln$ of both sides, we have
$\frac{0.56\,mg}{A}=e^{5.684}=294$
Or $A= \frac{0.56\,mg}{294}=0.0019\,mg $
