The liquid in each cone is itself a cone; with equal starting volumes, the narrow cone (radius $3$) is four times as tall as the wide cone (radius $6$)

Adding the same marble volume scales the height by the same factor $\sqrt[3]{1 + \frac{\Delta V}{V}}$ in both cones, so the rises are in the ratio of the original heights

The liquid in each cone forms a smaller cone similar to the container. Let the narrow liquid cone have radius $3$ and height $h_1,$ and the wide one radius $6$ and height $h_2.$ Equal volumes give $\tfrac13\pi\cdot 9\cdot h_1 = \tfrac13\pi\cdot 36\cdot h_2,$ so $h_1 = 4h_2.$

Dropping the marble raises the volume by the same amount $\Delta V = \tfrac43\pi$ in each cone, and both start with the same volume $V.$ Because a cone’s volume scales as the cube of its height, the new height is $h\sqrt[3]{1 + \frac{\Delta V}{V}},$ so each rise equals $h\left(\sqrt[3]{1 + \frac{\Delta V}{V}} - 1\right).$ This factor is identical for the two cones, so the rises are in the ratio $h_1 : h_2 = 4 : 1.$

Thus, the correct answer is E.