Answer:
[tex]\boxed{\text{Replacing the cube of manganese(IV) oxide with its powder}}[/tex]
Explanation:
The MnO₂ is acting as a catalyst: it takes part in the reaction but can be recovered at the end.
A. Using a 2 g cube of MnO₂
The surface area of the catalyst would double, so the rate would probably double. Technically, this option is correct, but I don't think it is the one the questioner intended.
B. Using 50 cm³ of H₂O₂
No effect. The rate depends on how fast the H₂O₂ molecules can reach the surface of the catalyst, and the surface area hasn't changed.
C. Using powdered MnO₂
This would increase the rate enormously, because the powdered catalyst has a much greater surface.
D. Removing MnO₂
The rate would decrease to near-zero, because there is no catalyst.