Solution:

The eight centers are at height $100,$ at the vertices of a regular octagon of side $200$ (adjacent spheres are tangent). If the ninth sphere has radius $r,$ it rests on the surface with its center at height $r$ directly above the octagon's center, and tangency to each sphere gives $\sqrt{R^2 + (r - 100)^2} = r + 100,$ where $R$ is the octagon's circumradius. Hence $$R^2 = (r + 100)^2 - (r - 100)^2 = 400r.$$

A side of a regular octagon subtends $45^\circ$ at the center, so $200 = 2R \sin 22.5^\circ$ and, using $\sin^2 22.5^\circ = \frac{1 - \cos 45^\circ}{2} = \frac{2 - \sqrt{2}}{4},$ $$R^2 = \frac{10000}{\sin^2 22.5^\circ} = \frac{40000}{2 - \sqrt{2}} = 20000\,(2 + \sqrt{2}).$$

Then $r = \frac{R^2}{400} = 50\,(2 + \sqrt{2}) = 100 + 50\sqrt{2},$ so $a + b + c = 100 + 50 + 2 = 152.$