Question 831364
{{{R}}}=radius of .
Since the ball fits tightly in the cube shaped box, the inside of the box must be {{{2R}}} wide.
Volume of the ball ={{{(4/3)pi*R^3}}}
Volume inside the box ={{{(2R)^3=8R^3}}}
The volume that needs to be filled with packing material is
{{{8R^3-(4/3)pi*R^3=(8-4pi/3)R^3}}}
As a fraction of the volume inside the box that is
{{{(8-4pi/3)R^3/8R^3=(8-4pi/3)/8=1-pi/6}}}
That is approximately {{{0.4764}}} , which is 47.64%.
 
It does not matter what size ball you fit tightly into a cube box, about 48% of the box will be empty space.