China Mathematical Competition (Shaanxi)

As in the diagram, let $A$, $B$, $C$, $D$ be the centers of the $4$ balls in the bottom layer, and $A^{\prime }$, $B^{\prime }$, $C^{\prime }$, $D^{\prime }$ the centers of the $4$ balls in the upper layer. Then $A$, $B$, $C$, $D$ and $A^{\prime }$, $B^{\prime }$, $C^{\prime }$, $D^{\prime }$ are the $4$ vertices of squares of length $2$, respectively. Now, the circumscribed circles with centers $O$ and $O^{\prime }$ of the squares constitute the bases of another cylinder, and the projecting point of $A^{\prime }$ on the bottom base is the middle point $M$ of arc $AB$.



In $△A^{\prime }{A}^{\prime }B$, we have $A^{\prime }A=A^{\prime }B=AB=2$, then $A^{\prime }N=\sqrt{3}$, where $N$ is the middle point of $AB$. Meanwhile, $OM=OA=\sqrt{2}$, $ON=1$, so $$MN=\sqrt{2}-1,A^{\prime }M=\sqrt{(A^{\prime }N{)}^2-(MN{)}^2}=\sqrt{8}.$$ Then the height of the original cylinder is $\sqrt{8}+2$.