Bounded Balls

This involves an application of Descartes' theorem . From equation (2) in the link, noting that the curvature $k$ of a circle of radius $r$ is given by $k = \dfrac{1}{r}$ , and taking the positive root to find the curvature $k_{4}$ of the inscribed circle, we have that

$k_{4} = \dfrac{1}{69} + \dfrac{1}{46} + \dfrac{1}{23} + 2\sqrt{\dfrac{1}{69*46} + \dfrac{1}{69*23} + \dfrac{1}{46*23}} =$

$\dfrac{1}{23}*\left(\dfrac{1}{3} + \dfrac{1}{2} + 1\right) + \dfrac{2}{23}\sqrt{\dfrac{1}{6} + \dfrac{1}{3} + \dfrac{1}{2}} = \dfrac{1}{23}*\dfrac{11}{6} + \dfrac{2}{23} = \dfrac{1}{23}*\dfrac{23}{6} = \dfrac{1}{6}$ .

Thus the radius of the fourth ball is $\dfrac{1}{k_{4}} = \boxed{6}$ .