Explanation:
that is a strange question. you will see in a moment why.
I suspect your teacher just wants to do the exercise to have a large volume and then see how often a small volume fits into that large volume.
we can do that quickly:
large volume = 10×2×312 = 6240 in³
small (cube) volume = 12×12×12 = 1728 in³
so, we could put 6240/1728 = 3.611111111... cubes into a volume as the prism has.
now why did I phrase this so strangely ?
because the prism not only has volume but also a shape.
it is a narrow but high container with the small dimensions being 10 in and 2 in.
the cubes have a side (or edge) length of 12 in. that is not compatible with the 10 in and 2 in cubes of the prism.
not a single one cube would fit into the prism.
so, the realistic answer is : 0 cubes.
if it does not matter that the cubes would stand out of the prism, we could simply stack a single file of cubes along the height of the prism :
312/12 = 26 cubes.
so, we could stack 26 cubes inside the height of the prism.
so, I conclude, this is either a trick question (and 0 is the true answer), or there is a mistake in the question.
possibly in the dimensions of either the prism or the cubes.
for example, let's assume the cubes have an actual side length of 2 in (and not 12 in).
then the small (cube) volume is : 2×2×2 = 8 in³
and they will fit into the prism also shape wise.
6240/8 = 780 cubes
so, we would need 780 cubes of that size to fill the prism.
or, as another example, the cubes have a side length of 1/2 in.
then the small (cube) volume = 1/2³ = 1/8 in³
6240/(1/8) = 6240 × 8 = 49,920 cubes
in that case we would need 49,920 cubes to fill the prism.