Answer: 813

Solution:

The rope runs from its anchor $A$ at the base of the tower, hugs the wall up to a point $P,$ then goes straight to its end $Q,$ which is at height $4$ and at distance $8 + 4 = 12$ from the tower's axis. Unroll the cylinder's wall into a plane: a taut rope becomes a single straight segment of length $20$ rising $4$ feet, so its horizontal projection has length $\sqrt{20^2 - 4^2} = 8\sqrt{6},$ and every piece of the rope has the same ratio $\frac{20}{8\sqrt{6}} = \frac{5}{2\sqrt{6}}$ of length to horizontal projection.

Viewed from above, the free portion $PQ$ is tangent to the circle of radius $8$ at $P$ from a point at distance $12,$ so its horizontal projection has length $\sqrt{12^2 - 8^2} = 4\sqrt{5}.$ Therefore $$\begin{aligned} PQ &= \frac{5}{2\sqrt{6}} \cdot 4\sqrt{5} \\ &= \frac{10\sqrt{5}}{\sqrt{6}} \\ &= \frac{5\sqrt{30}}{3}. \end{aligned}$$

The rope touching the tower has length $20 - \frac{5\sqrt{30}}{3} = \frac{60 - \sqrt{750}}{3},$ and $c = 3$ is prime, so $a + b + c = 60 + 750 + 3 = 813.$