$0.048$ mg
Step 1: The amount of a substance remaining after $n$ half-lives is given by: \[ N = N_0 \times \left(\frac{1}{2}\right)^n \] where $N_0$ is the initial mass, and $n$ is the number of half-lives elapsed.
Step 2: Given: \[ N_0 = 12 { mg}, \quad T_{1/2} = 3.5 { days}, \quad t = 35 { days} \] \[ n = \frac{t}{T_{1/2}} = \frac{35}{3.5} = 10 \]
Step 3: Calculate remaining mass: \[ N = 12 \times \left(\frac{1}{2}\right)^{10} \]
Step 4: \[ \left(\frac{1}{2}\right)^{10} = \frac{1}{1024} \approx 0.00098 \] \[ N = 12 \times 0.00098 = 0.0118 \approx 0.012 { mg} \]
Step 5: Therefore, the correct answer is (B).