I think so, too. Let's reduce the length, width and height of that box to 80,50, 10 (only equivalent to a sphere).

My placement method is as follows (line length and column width):

Put 5 in the first column (because it is 50cm, of course, you can put 5).

Second column: put a ball tangent to both balls between two adjacent balls in the first column (that is, embedded between the two balls), so that four balls can be placed in the second column.

In this case, the width of the two volleyballs is not 20cm.

An equilateral triangle with a side length of 10cm and a distance of 5 * (3) 0.5 from the vertex to the bottom.

This is a little shorter than the original 20cm, and the specific value is: 10-5 * (3) 0.5.

In this case, you can put 9 columns. (1st column 5, 2nd column 4, 3rd column 5, 4th column 4, ..., 8th column 4, 9th column 5)

This is because 10-5 * (3) 0.5 is added to every two columns, so a * * * can be added by 8 *10-5 * (3) 0.5 =10.718cm.

There is only enough room for one column, so you can put nine columns.

So 4 1 (5*5+4*4) can be put at the bottom of the box.

The width is 50cm and the height is 80cm. Like the bottom surface, they are placed at 5, 4, 5, 4 ... respectively, which is equivalent to turning the bottom surface over 90 degrees.

There are five in the first column on the first floor, four in the first column on the second floor (this situation is equivalent to bargain hunting), five in the first column on the third floor, four in the first column on the fourth floor ... five in the first column on the ninth floor.

The first column that can hold 5 can hold 5*5+4*4=4 1 on this floor.

Anything in the first column that can hold 4 can hold 5*4+4*5=40 on this floor.

So a * * can be 4 1*5+40*4=365.