If I have $10$ pounds of stainless steel balls of $2$ inch, then together they have a (pack)volume. Will the (pack)volume change if I switch to $10$ pounds of stainless steel $4 $ inch balls?
(I think it will change because of the air between the balls is more with the $4$ inch balls, so it will be a bigger box I need, but not sure).
It depends how you pack. Let's assume that you have a very large number of balls, so that you can ignore the effects of the balls on the edge. If you pack them the same way, say simple cubic, then yes, you have bigger air gaps, but you have fewer of them. In fact, from geometrical considerations only, you can show that the same arrangement has the same fraction of filled and void. Look up packing fraction. See this link for example.