#### Answer

$\approx 92.1 s$

#### Work Step by Step

Given: When the voltage is 10% of its original value, that is, $V =0.1 V_0$
Now, we have $V(t)=V_0e^{-t/40}$ ...(1)
This implies that $0.1 V_0=V_0e^{-t/40} \implies t=-40 (\ln (0.1)) $
This implies that $t \approx 92.1 s$
Hence, $t \approx 92.1 s$