Answer
$\approx$ $2500$ years
Work Step by Step
Let $y(t)$ be the level of radioactivity
So
$y(t)$ = $y(0)e^{-kt}$
and $k$ is determined by using the half-life
$y(5730)$ = $\frac{1}{2}y(0)$
$y(0)e^{-5730k}$ = $\frac{1}{2}y(0)$
$e^{-5730k}$ = $\frac{1}{2}$
$k$ = $\frac{\ln2}{5730}$
If $74$% of the C remains, then we know that
$y(t)$ = $0.74y(0)$
$0.74$ = $e^{\frac{-t(\ln2)}{5730}}$
$t$ $\approx$ $2489$ years $\approx$ $2500$ years