Answer: D

Solution:

The six unequal sectors form a fixed cycle of $6$ distinguishable positions, so we want proper $3$-colorings of a $6$-cycle that use each color exactly twice. There are $(3-1)^6 + (3-1) = 66$ proper $3$-colorings altogether. Of these, $6$ use only two colors: choose the missing color in $3$ ways, then alternate the other two colors in $2$ ways. Another $36$ have color counts $(3,2,1)$: choose the color used three times in $3$ ways, choose one of the $2$ alternating sets of positions for it, choose the color used twice in $2$ ways, and choose its $2$ positions among the other $3$ in $3$ ways. The remaining $66 - 6 - 36 = 24$ use each color exactly twice. Therefore, the answer is D.