Answer
about $250.39\ days$
Work Step by Step
Step 1. Given $A_0=50\ mg, A=34\ mg, t=30\ days$, we have $34=50e^{-30k}$, thus $k=-\frac{1}{30}ln\frac{34}{50}$
Step 2. To reach $A=2\ mg$, we have $2=50e^{-kt}$ and $t=-\frac{1}{k}ln\frac{2}{50}=30\cdot \frac{ln\frac{2}{50}}{ln\frac{34}{50}}\approx250.39\ days$
