Deciding capacitor value to use in my circuit
You start with $$ Q = CV $$ and differentiate both sides, $$ \dfrac{dQ}{dt}= C\dfrac{dV}{dt} $$, which can be represented by $$ I = C \dfrac{dV}{dt}$$, rearranging variables you get $$ \dfrac{I}{\frac{\Delta{V}}{\Delta{t}}} = C $$

Using your values from above, $$ \dfrac{2 [A]}{\frac{(3.3 - 3.2)[V]}{477 [us]}} = 9.54 [mF] $$