The volumes of the following 6 bodies are equal: a regular tetrahedron, a cube, a regular octahedron, a sphere, a cylinder, and a cone, the cross-section of the latter two bodies along their axis is a square, respectively an equilateral triangle. Arrange the bodies in ascending order of their surface area.
For each body with a common volume $V$, we express the body's edge, height, radius, and from this, the surface area $F$. The length measure will always be a constant multiple of $V^{1 / 3}$, and $F$ will be a constant multiple $k$ of $V^{2 / 3}$, so the ratio $F / V^{2 / 3}=k$ will be a characteristic constant number for each body. Based on this, we will sort the bodies. However, it will be more convenient to calculate $k^{3}$.

For the cube, the edge $a=V^{1 / 3}$, from which $F=6 a^{2}=6 V^{2 / 3}$, so $k=F / V^{2 / 3}=6$, and $k^{3}=F^{3} / V^{2}=216$.

In the regular tetrahedron, the height $m$ and the radius $r$ of the circle circumscribed around the base are the legs of a right triangle whose hypotenuse is the edge $c$. The radius, however, is $2 / 3$ of the height of the equilateral triangle base, because the latter is also a median, and the center of the circle is also a centroid: $r=(2 / 3) \cdot c \sqrt{3} / 2=c / \sqrt{3}, m=c \sqrt{2 / 3}, V=(1 / 3) \cdot c^{2}(\sqrt{3 / 4}) \cdot (c \sqrt{2 / 3})=c^{3} \sqrt{2} / 12, V^{2}=c^{6} / 72$, on the other hand $F=4 \cdot\left(c^{2} \sqrt{3} / 4\right)=c^{2} \sqrt{3}, F^{3}=3 \sqrt{3} c^{6}$, and thus $F^{3} / V^{2}=216 \sqrt{3} \approx 374.1$.

The regular octahedron can be cut into two equal-edged square-based pyramids by a plane through two opposite edges. In these, the height is half the diagonal of the base, because a similar square section can also be drawn through the vertex opposite the base. Denoting the edge by $d$, $V=2 \cdot\left(d^{2} / 3\right) \cdot(d / \sqrt{2}), V^{2}=2 d^{6} / 9$, on the other hand $F=8 \cdot\left(d^{2} \sqrt{3 / 4}\right), F^{3}=24 \sqrt{3} d^{6}$, so $F^{3} / V^{2}=108 \sqrt{3} \approx 187.1$.

For the sphere, $F^{3} / V^{2}=\left(64 \pi^{3} r^{6}\right) /\left(16 \pi^{2} r^{6} / 9\right)=36 \pi \approx 113.1$.

In the cylinder, $m=2 r$, so $V=2 \pi r^{3}, F=2 \pi r^{2}+2 \pi r \cdot m=6 \pi r^{2}$, hence $F^{3} / V^{2}=54 \pi \approx 169.6$.

Finally, for the cone, $m=r \sqrt{3}$, the slant height is $2 r$, so $V=\pi r^{3} / \sqrt{3}, F=\pi r^{2}+2 \pi r^{2}=3 \pi r^{2}, F^{3} / V^{2}=81 \pi \approx 254.5$.

Based on the $k^{3}$ values found for each body, the desired order is: sphere, cylinder, octahedron, cube, cone, tetrahedron.