Net moles of solute(HCL in this question) before and after dilution is same
hence MnetVnet = M1V1 + M2V2
V = (M1V1+M2V2)/Mnet
Put the values given in question to get
V net= 1200 ml
now total volume is 1200 ml and intial volume was 1000 ml
hence VOlume of water added is 200 ml