Using exponential decay formula:
[tex]N=N_o(\frac{1}{2})^{\frac{t}{\tau}}[/tex]
Where:
[tex]\begin{gathered} N_o=3mg \\ t=222.5_{\text{ }}days \\ \tau=44.5_{\text{ }}days \end{gathered}[/tex]
Therefore:
[tex]\begin{gathered} N=3(\frac{1}{2})^{\frac{222.5}{44.5}} \\ N\approx0.09mg \end{gathered}[/tex]
Answer:
About 0.09 miligrams
