Answer
$t=\lambda\dfrac{\ln (10)}{\ln 2}=19,035$ years
Work Step by Step
The decay equation can be written as:
$A=A_0e^{-kt}$
Half-life: $\lambda=5730$ years
Also, $k=\dfrac{\ln 2}{\lambda}\approx 1.20968 \times 10^{-4}$
Now, we have:
$0.1 A=A_0e^{-kt}$
This implies $t=\dfrac{-\ln (10)}{-k}$
Thus, $t=\lambda\dfrac{\ln (10)}{\ln 2}=19,035$ years