Answer
Within 4.9 seconds
Work Step by Step
Decay constant $k=\frac{0.693}{t_{1/2}}=\frac{0.693}{32\,s}=0.021656\,s^{-1}$
If original activity $A_{0}=100$
and activity at a later time $t$ is $A=90$, then
Using the reation $\ln(\frac{A_{0}}{A})=kt$, we get
$\ln(\frac{100}{90})=0.10536=0.021656\,s^{-1}(t)$
$\implies t=\frac{0.10536}{0.021656\,s^{-1}}=4.9\,s$
