Answer: 53

Small Hint:

Project the cut face straight down onto the circular face containing $A$ and $B:$ the image is the disk minus two $120^\circ$ circular segments

Big Hint:

With $O$ the cylinder’s center, $O'$ the face center, and $M$ the midpoint of $AB,$ the tilt satisfies $\cos\theta = \frac{O'M}{OM};$ divide the projected area by $\cos\theta$

Solution:

Stand the block on the face containing $A$ and $B,$ and let $O'$ be the center of that face, $M$ the midpoint of $\overline{AB},$ and $O$ the center of the cylinder. The cutting plane meets the bottom face in chord $\overline{AB}$ and, by symmetry through $O,$ meets the top face in the reflected chord, so the cut face projects vertically onto the region $R'$ between chord $\overline{AB}$ and its mirror image through $O'$ (shaded below). Each $120^\circ$ circular segment cut off has area $\frac{1}{3}\pi \cdot 6^2 - \frac{1}{2} \cdot 6 \cdot 6 \sin 120^\circ$ $= 12\pi - 9\sqrt{3},$ so $R'$ has area $36\pi - 2\left(12\pi - 9\sqrt{3}\right)$ $= 12\pi + 18\sqrt{3}.$

Since $\overset{\frown}{AB} = 120^\circ,$ triangle $AO'B$ gives $O'M = 6\cos 60^\circ = 3,$ and $OO' = 4,$ so $OM = 5.$ The cut face is planar and tilted from the horizontal only in the direction of $\overline{O'M},$ at the angle $\theta$ with $\cos\theta = \frac{O'M}{OM} = \frac{3}{5}.$ Undoing the projection therefore multiplies areas by $\frac{5}{3},$ so the unpainted face has area $\frac{5}{3}\left(12\pi + 18\sqrt{3}\right) = 20\pi + 30\sqrt{3}.$ Thus $a + b + c = 20 + 30 + 3 = 53.$