The highlighted sector is exactly 3/4 of the circle area, because [tex] \frac{3\pi}{2} [/tex] is 3/4 of [tex] 2\pi [/tex] which is a complete rotation.
Since you are given the radius, you can compute the area as
[tex] A = \pi r^2 = 16\pi [/tex]
So, 3/4 of this area would be [tex] 12\pi [/tex] squared cm.
The remaining sector, of course, is [tex] 4\pi [/tex] squared cm.
