Answer
$9.5\times10^{-4}\,mg$
Work Step by Step
Amount of radionuclide at the beginning $A_{0}=0.015\, mg$
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{78.25\,h}=0.008856\,h^{-1}$
Time $t=13\times24\,h=312\,h$
$\ln(\frac{A_{0}}{A})=kt$ where $A$ is the amount of radionuclide remaining.
$\implies \ln(\frac{0.015\,mg}{A})=0.008856\times312=2.763$
Taking the inverse $\ln$ of both sides, we have
$\frac{0.015\,mg}{A}=e^{2.763}=15.847$
Or $A= \frac{0.015\,mg}{15.847}=9.5\times10^{-4}\,mg$
