Solution:
Answer: $\sqrt{\frac{1}{3}}$

It can be seen that the points of tangency of the spheres with $\mathcal{E}$ must lie on its major axis due to symmetry. Hence, we consider the two-dimensional cross-section with plane $ABC$. Then the two spheres become the incentre and the excentre of the triangle $ABC$, and we are looking for the ratio of the inradius to the exradius. Let $s$, $r$, $r_{a}$ denote the semiperimeter, inradius, and exradius (opposite to $A$) of the triangle $ABC$. We know that the area of $ABC$ can be expressed as both $rs$ and $r_{a}(s - |BC|)$, and so $\frac{r}{r_{a}} = \frac{s - |BC|}{s}$. For the given triangle, $s = 6$ and $a = 4$, so the required ratio is $\frac{1}{3}$.