Answer
$1.1\times10^{6}\,y$
Work Step by Step
Number of moles $n=\frac{\text{mass in grams}}{\text{molar mass}}=\frac{1.0\times10^{-3}\,g}{79\,g/mol}=1.2658\times10^{-5}\,mol$
Number of particles $N=n\times\text{Avogadro number}$
$=1.2658\times10^{-5}\times6.022\times10^{23}=7.62265\times10^{18}$
Decay rate= $1.5\times10^{5}\,dps=kN=k\times7.62265\times10^{18}$
$\implies k=\frac{1.5\times10^{5}/s}{7.62265\times10^{18}}=1.9678\times10^{-14}\,s^{-1}$
$t_{1/2}=\frac{0.693}{k}=\frac{0.693}{1.9678\times10^{-14}\,s^{-1}}=3.5217\times10^{13}\,s$
$=3.5217\times10^{13}\,s\times\frac{1\,h}{3600\,s}\times\frac{1\,d}{24\,h}\times\frac{1\,y}{365\,d}=1.1\times10^{6}\,y$
