The water doesn't interfere with the acid or alkali, it only increases the total volume of the solution, thus diluting it making the solution less concentrated.
So as a general rule here, you calculate which is in excess, the acid or the alkali in moles. Then you calculate the concentration of [H+] from this, taking into the account the increase in total volume due to adding 55ml of water. Then we can find PH.
So to begin:
Moles of H+ in H2SO4=CxVx2= 0.001x(25x10^-3)x2=0.00005moles of H+
Moles of OH- in NaOH=CxV= 0.001x(20x10^-3)= 0.00002moles of OH-
Moles of H+ in excess = 0.00005-0.00002=0.00003
Where V=25+20+55=100ml
Therefore, [H+]=Mol/V= 0.00003/(100x10^-3) = 0.0003moldm^-3
So, PH=-log[H+]
PH=-log(0.0003)
PH= 3.52 (2dp)