For Quantity A, the vertical sides are obviously 4 and 12. Less obviously, the shaded region can be broken up into a square with a right isosceles triangle on the top and bottom. (The picture isn't drawn to scale at all.) The diagonals of the right triangles would thus be 4√2. Thus, the perimeter is 16 + 8√2.

For Quantity B, we can find it in two ways. Firstly, we can use the 3 shapes that we just mentioned. The square is 4x4=16 and the two right isosceles triangles can be shoved together to make another 4x4 square, making a total of 32.

Another way to find Quantity B would be to find the area of the big square, 12x12 = 144, and the area of the little square, 4x4=16, and subtract them to find the area of the outer ring. 144-16=128. The shaded area is one fourth of that, so 128/4=32 as well.

Finally, you should know that √2=1.4ish, or at least that it's smaller than 2, so 16 + 8√2 must be smaller than 32 and thus the answer is B.

(Note: the statement that the squares have the same center is very important in this problem. It's easy to assume that they have the same center when you draw your own version, but if we weren't told this, the problem becomes quite different. Be very careful with any geometry picture which appears to have symmetry!)