Answer
$\approx 81,100$ years
Work Step by Step
A decay model has the form $Q(t)=Q_{0}e^{-kt}$.
The decay constant k and half-life $t_{h}$ are related by
$t_{h}k=\ln 2$
Substitute $t_{h}=24,400$ and solve for k.
$24,400k=\ln 2$
$k=\displaystyle \frac{\ln 2}{24,400}\approx 0.00002841$
So,
$Q(t)\approx Q_{0}e^{(-0.00002841)t}$
Now, $Q_{0}=10$ g, $Q(t)=1$ g. Substituting, we solve for t.
$1=10e^{(-0.00002841)t} \qquad/\div 10$
$e^{(-0.00002841)t}=0.1\qquad/\ln(..)$
$-0.00002841t=\ln(0.1)$
$ t=\displaystyle \frac{\ln 0.1}{-0.00002841}\qquad$ ... round to 3 sig.dig.
$\approx 81,100$ years
