If you were to take an arbitrary 3-dimensional shape with finite surface area, then look at the volume of that shape, turn the volume into a long cylindrical string bunched up ideally inside the shape with the space between the string be the limit approaching zero, so there is a string inside a shape that fits exactly in the volume. What if you pull that string out of the shape and wrap it around the object, or more generally, what is the ratio of the string's surface area (unbundled)to the surface area of the original shape.
So is this ratio the same number for the volume and surface area of any shape?
Note that the units don't work-the volume of a shape is length$^3$ while the surface area is length$^2$. Let us apply that to a sphere of radius $R$. The volume is $\frac 43 \pi R^3$. If we fill that with a string of radius $r$, cross sectional area $\pi r^2$, the length is $L=\frac {\frac 43 \pi R^3}{\pi r^2}=\frac {4R^3}{3 r^2}$. The surface area it will cover is then $2rL=\frac {8R^3}{3r}$ (notice this is not the surface area of the string, it is the cross sectional area), so it will cover the $4 \pi R^2$ surface of the sphere $\frac {2R}{3r}$ times. As $r \to 0$, this goes to $\infty$. This will be general-shrinking the radius of the string will cover the outer shape more and more times.