12. E.

Let the rise height in the narrow container be $h_{1}$, and the rise height in the wide container be $h_{2}$. Then
$$
\begin{array}{l}
\frac{1}{3} \pi r_{1}^{2} h_{1}=V_{\text {ball }}=\frac{1}{3} \pi r_{2}^{2} h_{2} \\
\Rightarrow \frac{h_{1}}{h_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}}=\frac{36}{9}=\frac{4}{1} .
\end{array}
$$
12. As shown in Figure 1, two conical containers with their vertices pointing downwards are filled with solutions of the same mass. The radii of the liquid surfaces are $3$ and $6$, respectively. A small ball with a radius of 1 is placed in each container, and the balls are completely submerged in the liquid and sink to the bottom without any liquid spilling out. The ratio of the increase in liquid height in the narrow-mouthed container to that in the wide-mouthed container is ( ).
(A) $1: 1$
(B) $47: 43$
(C) $2: 1$
(D) $40: 13$
(E) $4: 1$