Below is a top view of the bases of the three cones, where $A$ is the center of one of the circular bases with radius $50$ , $B$ is the center between all three cones, $D$ and $C$ are intersection points of the segments and circle $A$ , and $x$ is the length between $B$ and $D$ .

Since the triangle formed by the three circular bases is an equilateral triangle, $\triangle ABC$ is a $30$ - $60$ - $90$ triangle, and since $AC$ is $50$ , $AB$ is $\frac{100\sqrt{3}}{3}$ , and since $AD$ is $50$ , $BD = x = \frac{100\sqrt{3}}{3} - 50$ .

Now this is a side view of one of the cone and sphere, where $A$ , $B$ , $D$ , and $x$ are the same as above, $E$ is the intersection point of the extension of the side of the cone and the line through the sphere perpendicular to the floor, $F$ is the point of tangency between the sphere and the side of the cone, $G$ is the top of the cone, and $r$ is the radius of the sphere:

Then $GD = \sqrt{50^2 + 120^2} = 130$ by Pythagorean's Theorem, and $\triangle ADG \sim \triangle BDE \sim \triangle FOE$ by angle-angle similarity.

Since $\triangle ADG \sim \triangle BDE$ , $\frac{BE}{120} = \frac{x}{50}$ , so $BE = \frac{12}{5}x$ .

Since $\triangle ADG \sim \triangle FOE$ , $\frac{OE}{130} = \frac{r}{50}$ , and since $OE = OB + BE$ , $\frac{\frac{12}{5}x + 120 - r}{130} = \frac{r}{50}$ , so $r = \frac{2x + 100}{3} = \frac{2(\frac{100\sqrt{3}}{3} - 50) + 100}{3} = \frac{200\sqrt{3}}{9} \approx \boxed{38.5}$

Addition: When viewing the cones and the sphere from the side, such that the bases of the two cones (seen as isosceles triangles) are just touching each other in the middle, some people are mistakenly assuming that the sphere is tangent to both isosceles triangles, like in Picture #1. However, since the sphere sits in the middle of three cones, the side view looks more like Picture #2, where the sphere is not tangent to both isosceles triangles but overlaps behind the isosceles triangles. The correct side view that shows the sphere tangent to one of the isoscles triangles is in Picture #3, where you are viewing it in such a way that two of the cones are overlapping each other (in the picture, the two overlapping cones are on the left, and the tangent cone is on the right).

The three solid cones define an interior space in the shape of an inverted (virtual) cone, the same shape at the solid cones - just larger - in which the sphere is contained. The horizontal distance from the tip of a solid cone to the top of the sphere is 100√3/3, or 2√3/3 times the size of the solid cones.

The 2-dimensional representation of this cone and sphere is an (inverted) isosceles triangle with inscribed circle. The dimensions of the triangle are larger than the solid cones by the scaling factor shown above: 200√3/3 base and 260√3/3 equal legs.

Use the formula for the radius of a circle inscribed in an isosceles triangle to obtain the answer of 200√3/9 ≈ 38.049.