Answer
4300 Bq
Work Step by Step
Number of moles n=$\frac{140\,g}{40\,g/mol}=3.5\,mol$
Number of particles N=$n\times N_{A}$
$=3.5\times6.022\times10^{23}=2.1077\times10^{24}$
Among these, only 0.0117% is radioactive.
Therefore,
$N=2.1077\times10^{24}\times0.000117$
$=2.466\times10^{20}$
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{1.26\times10^{9}\times365\times24\times3600\,s}$
$=1.744\times10^{-17}\,s^{-1}$
Radioactivity=$kN=1.744\times10^{-17}\,s^{-1}\times2.466\times10^{20}$
$=4300\,decays/s=4300\,Bq$