Answer
$2.53\times10^{12}\,s$
Work Step by Step
If original activity $A_{0}=100$, then present activity $A=10.0$.
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{24100\,y}$
$\ln(\frac{A_{0}}{A})=kt$
$\implies t=\frac{\ln(\frac{A_{0}}{A})}{k}$
$=\frac{\ln(\frac{100}{10.0})}{\frac{0.693}{24100\,y}}=8.00755\times10^{4}\,y$
$=8.00755\times10^{4}\,y\times\frac{365\,d}{1\,y}\times\frac{24\,h}{1\,d}=\frac{3600\,s}{1\,h}$
$=2.53\times10^{12}\,s$
