Falling circle

The cubic equation has a local minimum at $(0,0)$ . The largest circle will balance on this point and be tangent to the cubic at one other point (purple).

Let this other point be $B=(b,\frac{b^{3}}{3}-\frac{b^{2}}{2})$ and the slope at this point is the derivative $b^{2}-b$ , the perpendicular slope is $\frac{1}{b-b^{2}}$

The perpendicular line at this point is $y-(\frac{b^{3}}{3}-\frac{b^{2}}{2})=\frac{1}{b-b^{2}}(x-b)$

which has y-intercept $\frac{-1}{1-b}+\frac{b^{3}}{3}-\frac{b^{2}}{2}$

The distance from $B$ to this y-intercept must equal the y-intercept because both are the radius of the circle.

$\sqrt{b^{2}+\frac{1}{(1-b)^{2}}}=\frac{-1}{1-b}+\frac{b^{3}}{3}-\frac{b^{2}}{2}$

This actually becomes a 10th-degree polynomial, the root we want comes from a cubic factor. But solving numerically, the solution of interest is $b=2.4294445...$

Plugging this into the y-intercept above gives the solution: $\boxed{2.52816185205477778...}$