jmc

Let the height of the smallest cone (the one on top) be $h$ and let the radius of the circular base of that cone be $r$. Consider the 4 cones in the diagram: the smallest one on top (cone A), the top 2 pieces (cone B), the top 3 pieces (cone C), and all 4 pieces together (cone D). Because each piece of the large cone has the same height as the smallest cone and the same angle and vertex at the top, each of the 4 cones is a dilation of the smaller cone at the top. In other words, all four cones are similar. Because cone B has height twice that of cone A, its circular base has twice the radius as that of A. Likewise, cone C has three times the height, and thus 3 times the radius, and cone D has 4 times the height and 4 times the radius. Thus, using the formula for the volume of a cone, we get $$\begin{array}{rl}{V}_B& =\frac{1}{3}\pi (2r{)}^2(2h)=\frac{8}{3}\pi r^{2}h\\ V_C& =\frac{1}{3}\pi (3r{)}^2(3h)=\frac{27}{3}\pi r^{2}h\\ V_D& =\frac{1}{3}\pi (4r{)}^2(4h)=\frac{64}{3}\pi r^{2}h\end{array}$$Looking at the diagram, we can see that the largest piece will be the volume of cone D minus that of cone C: $$V_1=\frac{64}{3}\pi r^{2}h-\frac{27}{3}\pi r^{2}h=\frac{64-27}{3}\pi r^{2}h=\frac{37}{3}\pi r^{2}h.$$Also notice that the volume of the second largest piece is the volume of cone C minus that of cone B: $$V_2=\frac{27}{3}\pi r^{2}h-\frac{8}{3}\pi r^{2}h=\frac{27-8}{3}\pi r^{2}h=\frac{19}{3}\pi r^{2}h.$$Thus, the ratio of the volume of the second largest piece to that of the largest piece is $$\begin{array}{r}\frac{V_2}{V_1}=\frac{\frac{19}{3}\pi r^{2}h}{\frac{37}{3}\pi r^{2}h}=\frac{\frac{19}{\cancel{3}}\cancel{\pi }\cancel{r^2}\cancel{h}}{\frac{37}{\cancel{3}}\cancel{\pi }\cancel{r^2}\cancel{h}}=\boxed{\frac{19}{37}}.\end{array}$$