@sqdancefan could i, for this particular problem, make a whole cube and then delete that one part that isn’t apart of the whole? ahhh, i’m sooo confused.

You need to be careful to distinguish area from volume. You could delete a 4x4x5 cuboid if you were figuring volume, but it's different for area.

If you look at this figure from any direction in which the 5-inch length is left-to-right across your field of vision, you see only a face that is 8 inches by 5 inches. Those 4 faces add to 4 * (8 in)*(5 in) = 160 in^2.

If you look at the figure from the side, so you see an L shape, You can consider that to be an 8-inch square with a 4-inch square cut out of it. That area is 64 in^2 -16 in^2 = 48 in^2. (Alternatively, it is a 4x8 rectangle plus a 4x4 square. It still adds up to 48 in^2.) The sum of the areas of these left and right sides is then 2*48 in^2 = 96 in^2.

The total area is then 160 in^2 +96 in^2 = 256 in^2, the 4th selection.

The surface area of this design is 5x4=20x4=80. Then 8x4=32x4=128. Finally 128+80=208 in2

: the awnser would be c)1 2/3