Hi NandishSS,

While this question looks complex, it's actually built around some fairly simple Geometry. It might help to break this calculation down into 'pieces' and think about the rules involved in this 3-dimensional shape.

To start, the 'insert' is a rectangle that's been folded in 2 spots. By definition, it has a length and a width; once we figure out those two dimensions, we can figure out its area.

We're meant to assume that the insert 'touches' the sides, so the width of the insert has to match the width of the box. Since the base of the box is an 8-inch square, the width of the insert is also 8 inches.

Next, we'll work on the 'middle' horizontal piece (it's the easiest part) - since it's horizontal, then it has the same length as the box (which is also 8 inches). Thus, that 'middle piece' is an 8x8 square = 64 in^2

The two diagonal pieces are the same length, so once we figure out one, we can double it and get the total area of those 2 'pieces.' You should notice how a bunch of right triangles are formed. The base of each of those triangles is 8 and the height is 6. This is a classic 3/4/5 right triangle that's been doubled to become a 6/8/10. Thus, the two diagonal pieces are 8x10 rectangles = 80 in^2 each.

64+80+80 = 224

Final Answer: B

GMAT assassins aren't born, they're made,
Rich