Unlike the person above me, I'll try to help you. I haven't done the calculation to see if this is right, but here is how I would approach it:
First and foremost, understand that capacitance is how much ENERGY is stored on you capacitor. If you want more help with that,
http://en.wikipedia.org/wiki/Capacitance The minute you see energy and objects moving the first thing going in my head is always KE=PE at the end. Thus, kinetic energy gained by the electron as it accelerates is the energy provided by the capacitor (this is the easiest way for me to think about it, sorry). So, I'm guessing they told us the distance and time the e- took to travel so we can figure out the final velocity. Why do we need that? So we can plug it into KE=1/2mv^2.
So, you can use kinematics to get the final velocity of the electron as it hits the positive plate, right? Displacement=1/2at^2 so you can get acceleration, and then with that acceleration you can get the final velocity by doing finalvelocity^2=2(acceleration)(displacement)...in both cases, the Vinitial is disregarded since it's at rest. Now plug that into KE equation.
Now, that's our KE, it has to equal to PE. The PE for a capacitor is given by 1/2QV...when you rearrange it by plugging in V=Q/C (from Q=CV) for the V in 1/2QV, you get 1/2(Q^2)/C. So since all the potential energy of the capacitor was used to get to the positive plate, we can set KE=PE=0.5Q^2/C. Now, we know KE, we know Q, the charge of 2 coulombs, and all you have to do is solve for C and that should be the answer. Remember to watch your units when doing the kinematics and stuff. Let me know if this is not right!