Hey there!
I know this has been answered already, but any ideas on how to solve this? Wouldn't mind getting linked to the place to save someone the time of solving it (again) 
Saw this and thought I'd give it a try!
I'm assuming from my calculations that C is the correct answer... If not, let me know!
To start off with, if you have a solution with a pH of 1, then changing the pH to 2 will require a 10x decrease in the concentration of hydrogen ions, as the pH scale is a logarithm. As HCl is a strong acid and completely dissociates in water, you can assume that the 0.1 molarity given in the questions is also the hydrogen ion concentration.
Decreasing the hydrogen ion concentration by 10 would give it a value of 0.01 - you can check this using the pH=-log[H+] formula. Hence this is the molarity to aim for in the following calculation:
C1V1 = C2V2
0.1 x 0.09 = 0.01V2
0.009 = 0.01V2
V2 = 0.009 / 0.01
= 0.9 L or 900mL
Using the original concentration and volume, as well as the intended final concentration, 900mL is found to be the final volume of solution needed to obtain a pH of 2. Since the solution is already 90ml, then you can simply subtract 90 from 900 to give you the amount of water need for the dilution, which is: 900 - 90 = 810mL (option C).
Hope this helps!