jmc

In Crate A, we have 20 rows of 10 pipes packed directly on top of each other. So the height of the packing is 20 times the diameter of a single pipe, or 200 cm. In Crate B, draw a horizontal line through the centers of the 9 or 10 pipes in each row. By symmetry, the distance between each consecutive pair of these 21 lines will be the same, say equal to $d$. There will be 20 such distances.



The distance of the bottom line from the bottom of the crate is equal to the radius of a pipe, and the distance of the top line from the top of the top row is also equal to the radius of a pipe. Thus, the total height of the packing in Crate B is equal to $(10+20d)$ cm.

Next, we find $d$. If we extract three pairwise touching pipes from two consecutive rows, their centers form an equilateral triangle with side length equal to the diameter of each pipe, so $d$ is equal to the height of this equilateral triangle, i.e. $d=5\sqrt{3}$ cm. Therefore, the total height of this packing is $(10+100\sqrt{3})$ cm, which is approximately 183.2 cm.

Therefore, the difference in the total heights of the two packings is $$200-(10+100\sqrt{3})=\boxed{190-100\sqrt{3}}$$cm, or about 16.8 cm, with the packing in Crate A being the higher one.