Circumscribed Tetrahedron - 4 identical tangent spheres

The centers of the spheres form a regular tetrahedron $ABCD$ , with edge length $a=4$ . The tetrahedron that contains the four spheres is a dilation of $ABCD$ . The center of this dilation is the center $O$ of $ABCD$ . We just have to find the scale factor $r$ of the dilation.
In the figure, $OE$ is perpendicular to the base $ABC$ and extended meets the base of the image tetrahedron at $H$ . This extension has the length of the radius of the spheres, i.e. $EH=2$ Furthermore, $OE$ is $\dfrac{1}{4}$ the height of $ABCD$ , thus, $OE=\dfrac{1}{4}DE=\dfrac{1}{4}\times \sqrt{\dfrac{2}{3}}a=\dfrac{1}{4}\times \sqrt{\dfrac{2}{3}}\times 4=\sqrt{\dfrac{2}{3}}$ Hence, $r=\dfrac{OH}{OE}=\dfrac{OE+EH}{OE}=\dfrac{\sqrt{\dfrac{2}{3}}+2}{\sqrt{\dfrac{2}{3}}}=1+\sqrt{6}$ For the ratio of the volumes of the tetrahedrons we then have $\dfrac{{{V}_{{A}'{B}'{C}'{D}'}}}{{{V}_{ABCD}}}={{r}^{3}}\Rightarrow {{V}_{{A}'{B}'{C}'{D}'}}={{r}^{3}}\dfrac{{{a}^{3}}}{6\sqrt{2}}={{\left( 1+\sqrt{6} \right)}^{3}}\dfrac{{{4}^{3}}}{6\sqrt{2}}\approx \boxed{309.5838}$