jmc

Join the centres $A,$ $B,$ and $C$ of the three circles. The lines $AB,$ $BC,$ and $CA$ will pass through the points where the circles touch, so will each have length $10\text{ cm}$ (that is, twice the radius of one of the circles).

We can break the height of the pile into three pieces: the distance from the bottom of the pile to the line $BC,$ the height of the equilateral triangle $ABC,$ and the distance $A$ to the top of the pile.



The first and last of these distances are each equal to the radius of one of the circles, that is, $5\text{ cm}.$ So we must determine the height of $△ABC,$ which is an equilateral triangle with side length $10\text{ cm}.$ There are many ways to do this. Drop a perpendicular from $A$ to $P$ on $BC.$ Since $AB=AC,$ we know that $P$ is the midpoint of $BC,$ so $BP=5\text{ cm}.$



Then $△ABP$ is a $30^{\circ }$-$60^{\circ }$-$90^{\circ }$ triangle, so $AP=\sqrt{3}BP=5\sqrt{3}\text{ cm}.$ Thus, the height of the pile is $$5+5\sqrt{3}+5=\boxed{10+5\sqrt{3}}\text{ cm.}$$