Capacitor in series with voltage divider
The voltage in a capacitor in series with a resistor at time t can be determined from the following formula:
$$V_c=V_s (1-e^{-t/RC})$$
...where Vs is the voltage being applied to the capacitor, R is the resistance, C is the capacitance and t is the elapsed time.

However this only works for the characteristic circuit with a source, resistance and capacitance in series.

For your circuit we need to calculate the Thevenin source and resistance equivalent.  Luckily for the prior we have a simple voltage divider, so the new Vs will be 4.5V.  For the resistance we need to short circuit the voltage source and look at the resistance from the load terminals which will be R1 in parallel with R2.
$$R_{Th}=\frac{1}{\frac{1}{R1} + \frac{1}{R2}}$$
$$=500$$
So we now have the values to calculate the new charging characteristics.  A capacitor can be considered fully charged after 5 time constants:
$$\tau =RC$$
$$V_c=4.5\times(1-e^{-5})$$
$$=4.47V$$
We can see from above that the only way to reduce the charging time is to reduce the R or C values.  The C value remained the same, but the new R value was 500R instead of the original 800R and thus the new circuit will charge in 250ms instead of 400ms.