I thought this question was quite difficult relative to others.
First, because you have 1M of both NaOH and HCl, strong base and acid respectively, you will have to add 20mL to the NaOH to reach the equivalence point.
Then, you still have to add some more acid to get down to pH of 2. So you set it up using M1V1=M2V2.
(1M HCl)(x ml)= (10^-2 [H+] ( 40+x))
The 10^-2 is the concentration of H+ from pH 2 and this is tricky because you use 40+x because you are adding x amount of HCl on top of the 40ml you already have.
When you solve for x, you can neglect the x in 40+x becaues it will be really small so you get about .04.
So you will need to add a total of 20+.04 ml to get to pH=2 so the answer should be approx 20.04
