Charge flowing through switch when it is closed 
The problem that I'm having is that the initial charges would have been 300 and 600 microcoulomb on the 3 and 6 microfarad capacitors respectively. But when the switch is closed there's some charge that flows through it(300 microcoulomb to be precise) but i can't get to this conclusion. Please help... 
 A: "the initial charges would have been 300 and 600 microcoulomb on the 3 and 6 microfarad capacitors respectively". Although we talk about the charges on capacitors, this is a loose way of speaking and one that lets you down with this sort of question. The point is that capacitors have equal and opposite charges on their plates. The overall charge is zero.
If the circuit had been set up with the switch open, then, additionally, for both the right hand and left hand pairs, the charge on the bottom plate of the top capacitor is equal and opposite to the charge on the top plate of the bottom capacitor, since these plates and the link between them form an 'island' that can't have acquired any net charge. Treating the left hand pair and the right hand pair as capacitors in series, you'll find that the charges on the plates of each capacitor are ±400 $\mu$C. Ask if you don't understand this.
When the switch is closed, each capacitor has 100 V across it (again, ask if not clear) and the 3$\mu$F and 6$\mu$F have different charges. By considering the 'island plates' (now joined by a bridge to the other island!) you should easily work out how much charge flows through the switch.
A: Your new drawing looks right to me.  You started with two separate paths, and correctly added the capacitances in series, then got the amount of charge on each plate of each capacitor.  It's the same, but backwards, along each path initially.  You also know that those central segments are isolated initially, which means charges can't flow onto or off of them.  You correctly showed this in your diagram, since each capacitor had 400μC, but opposite signs.
Now, the top two are connected in parallel, and basically act as one capacitor.  Similarly, the bottom two are connected in parallel and act as one capacitor.  You seem to have correctly added each pair's capacitance in parallel, and then added them in series to get the right individual charges on each capacitor.
The important part to remember is the sign of the charge on each capacitor.  On the left side, the bottom plate of the top capacitor has -600μC, while the top plate of the bottom capacitor has +300μC, giving us a net -300μC on the left side.  Similarly, the right side has net +300μC.  But the part of the circuit with the switch in it is still isolated, so charge can't come from outside; it can only move from one side to the other.  That's why current flowed through.
