Let's go through all of them and find the equivalent capacitance.

(A) This is just 

(B) There are two capacitors in series, so this is 

(C) These capacitors are in parallel, so

(D) These are a combination of series and parallel.  Two are in series and they are in parallel with a third,

So, ranking them we get

In this question, we're told the maximum amount of charge that a capacitor can hold at a given voltage. We're then asked to determine the capacitance. To do this, we'll need to use the expression for capacitance.

Plug in the values given to us in the question stem:

The fact that the system is still connected to the battery indicate a constant V so regardless what happens to the capacitor, V stays fixed.

Relevant equations:

Plug the second equation into the first:

Considering all the variables in the numerator are held fixed for this problem, we see that increasing D will decrease the charge stored on each plate. 

The charge has no where to go. Without the battery connected, the charge has no physical avenue on or off the plates. 

As D increases, C will decrease

Does the battery matter? No. Capacitance is a "geometric" quantity. This means that it can be determined solely by physical parameters like the area and separation distance. The charge on the plates, voltage difference, electric field and any other quantity you could think of does not influence the capacitance other than A or D. 

The voltage rise through the source must be the same as the drop through the capacitor.

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

Combine equations and find the electric field:

Convert mm to m and plugg in values:

Use the electric field in a capacitor equation:

Convert  to  and plug in values:

The magnitude of total charge on the positive plate is equal to the total charge on the negative plate, so to find the number of excess elections:

The voltage rise through the source must be the same as the drop through the capacitor.

The voltage drop across the capacitor is the equal to the electric field multiplied by the distance.

Combine equations and solve for the electric field:

Convert mm to m and plug in values:

Using the electric field in a capacitor equation:

Rearrange to solve for the charge:

Convert  to  and plug in values:

Voltage difference  is given by

, where  is the electric field strength and  is the distance between the two plates. 

For this problem,

The formula for capacitance  is given by:

,

Where  is dielectric strength,  is distance between plates,  is permittivity of empty space, and  is cross sectional area. 

To determine , we do

 because between the plates is a vacuum.

Putting it all together,