Answer
$7.1\times10^{10}\,Bq$
Work Step by Step
$T_{\frac{1}{2}}= 1.23\times10^{6}\,s$
Number of atoms= $\frac{6.7\times10^{-6}g}{32g/6.022\times10^{23}}=1.26\times10^{17}\,atoms$
The decay rate R= $\lambda N= \frac{0.693}{T_{\frac{1}{2}}}N= $
$\frac{0.693\times1.26\times10^{17}}{1.23\times10^{6}s}=7.1\times10^{10}\,Bq$
