Answer
${\bf 0.69}\;\rm ms$
Work Step by Step
First of all, we need to find the time constant of this circuit which is given by
$$\tau=R_{eq}C_{eq}$$
So we had to reduce this circuit, as shown below, to one equivalent capacitor with one equivalent resistor.
Plug the known,
$$\tau=R_{eq}C_{eq}=(50\mu)(20)=\bf1.0\;\rm ms$$
As we see in the circuits below, the current $I$ that passes through the $8\;\Omega$ resistor is the full current from the equivalent capacitors which is the same current that passes through the equivalent resistor of $20\;\Omega$.
So, the discharge of the equivalent capacitor
$$I=I_0e^{-t/\tau}$$
when $I=0.5I_0$;
$$\frac{1}{2} \color{red}{\bf\not} I_0= \color{red}{\bf\not} I_0e^{-t/\tau}$$
Thus,
$$\dfrac{-t}{\tau}=\ln\left[ \frac{1}{2} \right]$$
$$t=-\tau\ln\left[ \frac{1}{2} \right]$$
Plug the known;
$$t=-(1\;\rm m)\ln\left[ \frac{1}{2} \right]$$
$$t=\color{red}{\bf 0.69}\;\rm ms$$
