Answer
$R = 3.66 \times 10^7 Bq $
Work Step by Step
The equation of rate of decay, $R = \lambda N$ and $\lambda = \frac{ln2}{T_{1/2}}$
$R = \frac{ln2}{T_{1/2}} N$
$R = (\frac{ln2}{(1600 y)(3.15 \times 10^7 s/y )}) ( \frac{(1.00 mg) (6.022 \times 10^{23}/mol )}{226 g/mol} )$
$R = 3.66 \times 10^7 s^{-1}$ or $3.66 \times 10^7 Bq $
