C₁V₁ = C₂V₂ should only be used to calculate the concentration change as a result of a volume change (dilution by water, basically).

Since the steps are actually more complicated than that (i.e.: there is a reaction involved), we have to consider this on from first principles, the basics:

1. 25.0 mL of 0.02 M KMnO₄ = 0.0005 mol of KMnO₄ was reacted away
2. Stoichometrically, this requires 5/2 x 0.0005 = 0.00125 mol of H₂O₂
3. We were told that the process took 20 mL of H₂O₂. Hence, c = n/V = 0.00125 / 0.020 = 0.0625 M

Umm.. there is just one hitch with the answer above, that is that the reaction took place in 45mL not the 20 mL of H2O2... and also another 8 moles of H2O was produced, so the correct concentration needs to include these both

So following the same basics above

1. 25.0mL of 0.02M KMnO4 = 25mL x 0.02M = 0.5 mmol
2. Stoichometrically n(H2O2) = 5/2 x 0.5 mmol = 1.25 mmol
3. Find volume of 8 moles of H2O V(H2O) = 8 x 18.02 x 1 = 144.16 mL
   --> assuming density of H2O to be 1 g/mL
4. Final volume of reaction = 45mL + 144.16mL = 189.16mL
5. Then final concentration of H2O2, c = n/V cfinal(H2O2) = 1.25mmol / 189.16 mL = 0.006608M
6. Round sig figs c(H2O2) = 6.61 x 10E-3 M

Note that this reaction would be exothermic so some of the water would be evaporated or hydrated the released oxygen gas so the concentration would be smaller than this theorectical one