What is the volume occupied by the atoms and the edge length in these cubic lattices?
This is the case of CCP (simple cubic closed packing) and FCC (face centered cubic closed packing)
For CCP
Imagine sitting in a cubical room. Now place some big spheres on each of the vertices of the room such that the centre of sphere and the vertices coincide.
You will see that some part of sphere goes outside the room itself. In chemistry, this is called sharing.
If your room is surrounded in all directions by other rooms, you share one sphere with 7 other rooms. (top 4 and bottom 4)
Thus a single particle gets shared by 8 rooms. The part in your room is 1/8
And the number of such particles are 8 (as cubic room has 8 vertices)
Net particles in your room $= \frac{1}{8} \cdot 8 = 1$
So in CCP the space occupied by all eight particles is basically equal to that of a single particle.
Now as assumed in the question, the particles are tightly packed, the
Radius of sphere equals half of the edge length. In other words edge length is twice the radius of the sphere.
$$\pu{Packing efficency = \displaystyle \frac{\text{Volume of sphere}}{\text{Volume of cube}}}$$
$$= z \cdot \frac{\frac{4}{3} \pi r^3}{(2r)^3} $$
(Here $z$ is net particle contribution; for the current example it is equal to 1)
Calculating this you'll get packing efficiency $\approx 52.4%$

For FCC
Now imagine putting one more sphere in the room, but this time, at the centers of the face. (Do this for each face). Now the particle in center gets shared only by 2 rooms. Yours and the room which is opposite to you.
Therefore, each particle has net contribution of half.
Since a cube has six faces, the net contribution of all spheres at the center the of sphere are
$$\frac{1}{2} \cdot 6 = 3$$
Add the already existent corner particles to it (whose net contribution is 1 as we know)
Thus the total number of particles in FCC are $3 + 1 = 4$

If you're good at imagining things you might have observed that now spheres won't fit in until you reduce their sizes. So we must falsify the assumption that twice the radius of sphere is the edge length. Spheres have shrunk.
I want you to imagine the face of the room now.
There is little space between two spheres on corner but not between the center sphere and corner spheres. Draw a diagonal line in the sphere passing through the diameter of central sphere. Now using Pythagorean theorem:
$$ \pu{(\text{Edge of cube})^2 + (\text{Edge of cube})^2 = (\text{Diagonal of cube})^2}$$
Let edge be 'A':
$$(A)^2 + (A)^2 = D^2$$
$$2(A)^2 = D^2$$
Since the center spheres have no space with corners ones. Therefore
4(radius of sphere) = Diagonal length
$$\sqrt{2}A = D$$
$$\sqrt{2}A = 4r$$
$$\sqrt{2}A / 4 = r$$

There you go. You've found the radius of sphere. Now put it in packing efficiency formula and take $z$ = 4
You will find it somewhere near 72%
On a sidenote: Observe that simple cubic has only 1 sphere which takes up nearly 50% of space of cube. While face centred has 6 + 8 = 14 smaller spheres which take up nearly 70 % of all space in cube. It's like comparing apples and bananas.
You might think that FCC is better than CCP as it uses space judiciously. But since in CCP, the particle size is large therefore CCP is best suited for particles of that size and FCC for smaller sizes.
Although we can discuss hexagonal packing form with CCP.