Video solution:

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Written solution:

Let the initial liquid heights in the narrow and wide cones be $h_1$ and $h_2$. Since the liquid volumes are equal,

$$\frac13\pi(3)^2h_1=\frac13\pi(6)^2h_2,$$

so $h_1=4h_2$.

After the identical marbles are dropped in, each cone must contain the same final volume below the liquid surface: the original liquid volume plus the volume of one marble. If the new liquid-surface radii are $3x$ and $6y$, similarity gives new heights $h_1x$ and $h_2y$. Thus

$$\frac13\pi(3x)^2h_1x=\frac13\pi(6y)^2h_2y.$$

Using $h_1=4h_2$, this simplifies to $x^3=y^3$, so $x=y$. The rise ratio is therefore

$$h_1(x-1):h_2(y-1)=h_1:h_2=4:1.$$

Thus, E is the correct answer.