For this problem, we are given the dimensions of a quarter and we need to determine the surface area of a roll of quarters.
We can approximate the roll as a cylinder, where the height is the sum of the heights of all the quarters and the dimater is equal to the diameter of one quarter. Therefore we have:
![\begin{gathered} A_(base)=\pi r^2=\pi((24.26)/(2))^2=462.24\text{ mm^^b2}\\ \\ h=40\cdot1.75=70\text{ mm}\\ \\ L_(base)=2\pi((24.26)/(2))=76.22\text{ mm}\\ \\ A_(lateral)=70\cdot76.22=5335.4\text{ mm^^b2}\\ \\ A_(surface)=2\cdot462.24+5335.4=6259.88\text{ mm^^b2} \end{gathered}]()
The surface area is equal to 6259.88 mm, the correct option is C.