In the figure above, square AFGE is inside square ABCD such

Dear PareshGmat,
That's a great question. I think it is very much like something the GMAT could ask on one of the harder questions.

Here's a solution.
If we take the big square ABCD (Area = 64) and subtract both the quarter circle and the little square AFGE, then divide the remaining area in half, we will get the shaded area. Very elegant.
Of course, big square ABCD = 64.
Quarter circle = \(16\pi\)
What about the little square?

Draw diagonal AC, which has a length of \(AC = 8sqrt(2)\).
Of course, GC is just the radius of the circle: GC = 8
Well, \(AG = AC - GC = 8sqrt(2) - 8 = 8(sqrt(2) - 1)\)
That's the diagonal of the square AFGE.

As a shortcut, we can find the area of any square by squaring the diagonal, and then dividing by two.
Area of AFGE = \(\frac{(AG)^2}{2}\)
Area of AFGE = \(\frac{(8(sqrt(2) - 1))^2}{2}\)
Area of AFGE = \(32(sqrt(2) - 1)^2\)
Area of AFGE = \(32(2 - 2sqrt(2) + 1)\)
Area of AFGE = \(96 - 64sqrt(2)\)

Twice shaded region = (Square ABCD) - (quarter circle) - (Square AFGE)
Twice shaded region = \(64 - 16\pi - (96 - 64sqrt(2))\)
Twice shaded region = \(64sqrt(2) - 32 - 16\pi\)

Divide by 2
Shaded region = \(32sqrt(2) - 16 - 8\pi\)
Factor out an 8
Shaded region = \(8(4sqrt(2) - 2 - \pi)\)
Shaded region = \(8(4sqrt(2) - (2 + \pi))\)
Answer = (B)

Great problem!
Mike