Answer
$\approx$ $12,457$ million years
Work Step by Step
$y(t)$ = $y(0)e^{-kt}$
$y(1250)$ = $\frac{1}{2}y(0)$
$y(0)e^{-1250k}$ = $\frac{1}{2}y(0)$
$e^{-1250k}$ = $\frac{1}{2}$
$k$ = $\frac{\ln2}{1250}$
$y(t)$ = $y(0)e^{-\frac{(\ln2)t}{1250}}$
$y(68)$ = $y(0)e^{-\frac{(\ln2)(68)}{1250}}$ = $0.963y(0)$
indicating that about 96% of the $40^K$ is remaining
$y(t)$ = $y(0)e^{-\frac{(\ln2)t}{1250}}$
$0.001y(0)$ = $y(0)e^{-\frac{(\ln2)t}{1250}}$
$0.001$ = $e^{-\frac{(\ln2)t}{1250}}$
$t$ $\approx$ $12,457$ million years