Answer
$$3.2 \times 10^{12} \mathrm{\ Bq}
$$
Work Step by Step
Equation $42-20$ leads to
$$
\begin{array}{l}{R=\frac{\ln 2}{T_{1 / 2}} N=\frac{\ln 2}{30.2 \mathrm{y}}\left(\frac{M_{\operatorname{san}}}{m_{\text {acom }}}\right)=\frac{\ln 2}{9.53 \times 10^{8} \mathrm{s}}\left(\frac{0.0010 \mathrm{kg}}{137 \times 1.661 \times 10^{-27} \mathrm{kg}}\right)} \\ {=3.2 \times 10^{12} \mathrm{Bq}}\end{array}
$$
