Answer
$0.17\,\mu g$
Work Step by Step
Amount of radionuclide at the beginning $A_{0}=2.8\mu g$
Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{2.69\,day}=0.2576/day$
Time $t=10.8\,day$
$\ln(\frac{A_{0}}{A})=kt$ where $A$ is the amount of radionuclide remaining.
$\implies \ln(\frac{2.8\,\mu g}{A})=0.2576\times10.8=2.782$
Taking the inverse $\ln$ of both sides, we have
$\frac{2.8\,\mu g}{A}=e^{2.782}=16.15$
Or $A= \frac{2.8\,\mu g}{16.15}=0.17\,\mu g$
