Answer
1.56 mg
Work Step by Step
The amount of radionuclide at the beginning is $A_{0}=25.0\,mg$
Rate constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{8.04\,day}=0.086194/day$
Time $t=32.2\,days$
$\ln(\frac{A_{0}}{A})=kt$ where $A$ is the amount of radionuclide remaining.
$\implies \ln(\frac{25.0\,mg}{A})=0.086194\times32.2=2.77545$
Taking the inverse $\ln$ of both sides, we have
$\frac{25.0\,mg}{A}=e^{2.77545}=16.046$
Or $A= \frac{25.0\,mg}{16.046}=1.56\,mg $
