Answer
$9.20s$
Work Step by Step
We can find the required time as follows:
$A=0.900A_{\circ}=A_{\circ}e^{\frac{-bt}{2m}}$
$\implies 0.900=e^{\frac{-bt}{2m}}$
We take the log on both sides to obtain:
$t=\frac{-2m}{b}(ln 0.900)$
We plug in the known values to obtain:
$t=-\frac{2(0.00144Kg)}{3.30\times 10^{-5}Kg/s}(\ln 0.900)$
$t=9.20s$
