A camera flash has a capacitor of 100 µF charged to 330 V…

Step 1. The charge. $Q = CV = 100 \times 10^{-6}\ \text{F} \times 330\ \text{V} = 3.3 \times 10^{-2}\ \text{C}$, that is $33$ mC.

Step 2. The current. $I = \dfrac{Q}{t} = \dfrac{3.3 \times 10^{-2}\ \text{C}}{2.0 \times 10^{-3}\ \text{s}} = 16.5\ \text{A}$.