Solution:
When the water in the cone is $h$ inches high, it forms a cone similar to the original, so that its base has radius $h / 4$ and its volume is hence $\pi h^{3} / 48$. The given condition then states that
$$\frac{d}{d t}\left(\frac{\pi h^{3}}{48}\right) = -h \Rightarrow \frac{\pi h^{2}}{16} \cdot \frac{d h}{d t} = -h \Rightarrow 2 h \cdot \frac{d h}{d t} = -\frac{32}{\pi} .$$
Integrating with respect to $t$, we get that $h^{2} = -\frac{32 t}{\pi} + C$; setting $t = 0$, $h = 12$, we get $C = 144$. The cone empties when $h = 0$, so $0 = -\frac{32 t}{\pi} + 144 \Rightarrow t = \frac{9 \pi}{2}$.