Let d = diameter of a golf ball.

Let us assume that the balls meet at full diameters---not at dimple-and-full-diameter---so that, say, any two adjacent balls is 2d "long".

Between a wall or floor of the box and a ball is assumed to be a full-diameter-and-wall contact---not by dimple-and-wall.

A rectangular box of 4d by 3d will fit the 12 balls. So will a 12d by d. So will a 6d by 2d.

Any of those 3 possible boxes has a volume V1 = 12d^3.

[4d*3d*d = 12d^3; 12d*d*d = 12d^3; 6d*2d*d = 12d^3]

Volume of 12 balls, V2 = 12[(4/3)(pi)(d/2)^3] = 2pi(d^3)

So, the "unfilled" space within the box is 12d^3 - 2pi(d^3) = (12 -2pi)(d^3).

The percentage of that on 12d^3 is

[12-2pi](d^3) / (12d^3)

= (12 -2pi)/12

= 0.4764

= 47.64 percent ----------answer.