By definition y vol #H_2O_2# means that sample solution of #H_2O_2# from each ml of which yml #O_2# at STP is produced on thermal decomposition.
So y ml #H_2O_2# solution will produce #y^2# ml #O_2# at STP.
Now the balanced equation of thermal decomposition reaction of #H_2O_2# is as follows
#2H_2O_2" "" "->" "" "O_2+2H_2O#
#=2" "mol" "" "=22400mL" ""at STP"#
This equation reveals that 22400mL #O_2# at STP is produced from 2mol #H_2O_2#
So #y^2# mL #O_2# will be produced from #(2y^2)/22400# mole #H_2O_2# and this amount of #H_2O_2# is present in #ymL# solution.
Again x mL y(M) #KMnO_4# solution will contain #(xyxx10^-3)# mole #KMnO_4#
Now the balanced redox reaction is
#5H_2O_2+2KMnO_4+3H_2SO_4->K_2SO_4+8H_2O+4O_2# comparing the mole ratio of reactants calculated in the reacting solutions with that of balanced redox reaction we can write
#(xyxx10^-3)/((2y^2)/22400)=2/5#
#=>x/y=0.4/11.2~~0.036#
