Answer
The time required is $~~32,000~years$
Work Step by Step
We can find $\lambda$:
$\lambda = \frac{ln(2)}{T_{1/2}}$
$\lambda = \frac{ln(2)}{5730~y}$
$\lambda = 0.000121~y^{-1}$
We can find the time required to reduce the activity to a fraction of $0.020$ of the initial activity:
$R = R_0~e^{-\lambda~t} = 0.020~R_0$
$e^{-\lambda~t} = 0.020$
$e^{\lambda~t} = 50$
$\lambda~t = ln(50)$
$t = \frac{ln(50)}{\lambda}$
$t = \frac{ln(50)}{0.000121~y^{-1}}$
$t = 32,000~y$
The time required is $~~32,000~years$
