How many ml of a 2 M solution of HCL would be needed to neutralize 2 ml of a 5 M solution of KOH?

You have .002 liter x 5 (mole/l) = 0.010 moles of base.

You will need an equal number of moles of acid
X (liters) * 2 mol/liter = 0.01 mole
X = 0.005 l = 5 ml