An air capacitor of capacitance C, plate area A and plate separation d has been filled with three dielectrics of constant $k_1=3,k_2=6and{k}_3=4$ , as shown in figure. The new capacitance XC, where X is?

Shortcut method: 

Before solving this problem try to understand the following concept. When a capacitor of capacitance C is divided into two parts in such a manner that each capacitor plate has area, half of the earlier value then each capacitor is understood to be connected in parallel and each one has the value C/2 as shown below:
 

But if you divided the gap of the capacitor in two equal parts parallel to the plates then each one will have the capacitance 2C (as ‘d’ has been halved) connected in series to each other as shown below:

Now in the given question you can see the capacitor has been divided into three chambers viz. left half and the right half which is further divided into two parts i.e. upper half and lower half.

So, all you have to do first is that write down the value of these separate chambers in terms of C. Left half is $\frac{\mathrm{C}}{2}$ and right half $\frac{\mathrm{C}}{2}$  is multiplied by 2 each for upper and lower half.

(That means right side C and C are in series and they are in parallel to C/2 making the entire capacitance C)
Now put the respective dielectrics in the chambers, so each chamber will have its new level of capacitance K (dielectric constant) times the old value.
 

Now the right side 6C and 3C and in series, so their equivalent value is $\frac{6C\times 3C}{6C+3C}=2C$  and this 2C is in parallel to the left side chamber capacitance 2C. So, the total capacitance becomes 4C.