Answer
$t = 2.08\times 10^{-4}~s$
Work Step by Step
The potential across the capacitor will be equal to the potential across the resistor when the potential across the capacitor is half of the battery emf.
We can find the time $t$:
$\frac{q}{C} = \mathscr{E}~(1-e^{-t/\tau}) = 0.500~\mathscr{E}$
$1-e^{-t/\tau} = 0.500$
$e^{-t/\tau} = 0.500$
$e^{t/\tau} = 2.00$
$\frac{t}{\tau} = ln(2.00)$
$t = ln(2.00)~\tau$
$t = ln(2.00)RC$
$t = [ln(2.00)]~(20.0~\Omega)~(15.0~\mu F)$
$t = 2.08\times 10^{-4}~s$
