Question 1064392
When packing spheres into a container,
there is always "wasted" empty space between the spheres.
There is also empty space between the spheres and the boundaries (floor, walls, ceiling) of the container.
The volume of a sphere with 6 inch diameter (3 inch radius) is
{{{Volume[ball]=(4/3)*pi*(3in)^3=36pi}}}{{{in^3=about113.1}}}{{{in^3}}} .
A cube with 1 foot edges has a volume of
{{{(12in)^3=1728}}}{{{in^3}}} .
With the packing you visualized first,
the fraction of the space filled with balls is
{{{8*36*pi/1728=about0.5236=52.36/100="52.36%"}}} .
You would fit only {{{8*960=7680}}} balls.
 
The alternate packing you propose is a better option,
especially for filling a large room with relatively small balls.

{{{drawing(500,300,-4.2,5.8,-3,3,
circle(0,0,1),circle(2,0,1),circle(-2,0,1),
circle(-3,-1.732,1),circle(-1,-1.732,1),
circle(1,-1.732,1),circle(3,-1.732,1),
circle(-1,1.732,1),circle(1,1.732,1),
green(arrow(-3.4,-1.7327,7,-1.732)),green(arrow(-2.4,0,7,0)),
line(-5,-2.732,6,-2.732),red(triangle(0,0,-1,-1.732,0,-1.732)),
red(triangle(0,0,-1,-1.732,1,-1.732)),red(rectangle(0,-1.732,0.2,-1.532)),
arrow(4.2,-1.732,4.2,0),arrow(4.2,0,4.2,-1.732),
locate(-0.6,-1.73,red(R)),locate(0.4,-1.73,red(R)),
locate(4.3,-0.7,sqrt(3)R)
)}}} The height used up for one layer of balls is {{{sqrt(3)R=about1.732R}}} rather than {{{2R}}} .
For a 1 cubic foot container, even with that strategy,
you cannot achieve a very efficient packing of balls
with a diameter of {{{6in=0.5ft}}} .
  
For the most efficient packing, the maximum theoretical fill ratio is,
as you found, {{{0.74=74%}}} .
You can approach that ratio, if the size of your container
is large compared to the volume of 1 sphere.
The size of your container is
{{{(10ft)*(12ft)*(8ft)=960}}}{{{ft^3=960}}}{{{ft^3*(1728in^3/ft^3)=1658880}}}{{{in^3}}} .
If you could fill 74% of that space with balls, you would fill
{{{0.74*1658880}}}{{{in^3=1227571.2}}}{{{in^3}}} .
The number of balls that fit in that "fillable" space is
{{{1227571.2/113.1=10854.145}}} (rounded).
So, {{{highlight(10854)}}} may be a good estimate.
 
For a more accurate calculation,
accounting for additional space wasted along floor, ceiling and walls,
we can calculates how many layers of balls fit into {{{8ft=96inches}}} ,
at {{{0.1732*3in=5.196in}}} per layer.
That would be {{{96/5.196=18.476}}} layers.
That is 18 layers, with some wasted space above the top layer.
Layers 1, 3, 5, etc would have {{{(20 balls)*(24balls)=480balls}}} .
Layers 2,,4,6,etc would have {{{(19balls)*(23balls)=437balls}}} .
with 18 layers, you would have
{{{9*(480balls)+9*(437balls)=highlight(8235balls)}}} .
That is better than {{{7680balls}}} in {{{16}}} layers of {{{480balls}}} per layer,
but it is far less than the {{{10854balls}}} estimate