Let A = the amount of contaminant in the water
$\displaystyle \frac{dA}{dt}=kA$
$\displaystyle \frac{dA}{A}=k\,dt$
$\displaystyle ln|A|=kt+C$
Let the amount of contaminant when t = 0 be 100
$\displaystyle ln100=C$
$\displaystyle ln|A|=kt+ln100$
Since 20% of the contaminants can be removed during the first minute
A=80 when t = 1
$\displaystyle ln80=k(1)+ln100$
$\displaystyle k=ln\frac{4}{5}$
$\displaystyle ln|A|=(ln\frac{4}{5})t+ln100$
You want to find t when A = 2