Answer
$r=(\sqrt 3-\dfrac{3}{2})m$
Work Step by Step
Suppose $1$ ball has center at $(r,r,r)$
$(x-r)^2+(y-r)^2+(z-r)^2=r^2$
$ \lt \frac{m}{2},\frac{m}{2},\frac{m}{2} \gt =(1+\frac{2}{\sqrt 3}) \cdot \lt r,r,r \gt$
$\frac{m}{2}=r(\frac{\sqrt 3+2}{\sqrt3})$
$r=\dfrac{\sqrt 3}{4+2\sqrt3}m$
Hence, $r=(\sqrt 3-\dfrac{3}{2})m$