Answer
$\tau = 2 \times 10^{-3}~s$
Work Step by Step
We can find the equivalent capacitance of $2~\mu F$ and $2~\mu F$ in series:
$\frac{1}{C} = \frac{1}{2~\mu F} + \frac{1}{2~\mu F}$
$\frac{1}{C} = \frac{2}{2~\mu F}$
$C = \frac{2~\mu F}{2}$
$C = 1~\mu F$
The equivalent resistance of $1~k \Omega$ and $1~k \Omega$ in series is $R = 2~k \Omega$
We can find the time constant $\tau$:
$\tau = R~C$
$\tau = (2 \times 10^3~\Omega)(1 \times 10^{-6}~F)$
$\tau = 2 \times 10^{-3}~s$
