Answer
64 hours.
Work Step by Step
Difference in time between 1.00 P.M. December 3, 2006 and 2:15 P.M. December 17, 2006 is 14 days, 1 hour and 15 minutes.
In minutes, $t= (14\times24\times60)+(1\times60)+ 15=20235\,min$
Original activity $A_{0}=9.8\times10^{5}\,dpm$
Current activity $A=2.6\times10^{4}\,dpm$
$\ln(\frac{A_{0}}{A})=kt$ where $k$ is the decay constant.
$\implies \ln(\frac{9.8\times10^{5}}{2.6\times10^{4}})=3.63=k(20235\,min)$
$\implies k=\frac{3.63}{20235\,min}=1.794\times10^{-4}\,min^{-1}$
Half-life $t_{1/2}=\frac{0.693}{k}=\frac{0.693}{1.794\times10^{-4}\,min^{-1}}$
$=3863\,min=\frac{3863}{60}\,h= 64\,h$
