Sveres

$Let\quad r\quad be\quad radius\quad of\quad the\quad spheres.\quad \quad For\quad a\quad sphere\quad tangent\quad \\ to\quad the\quad 3\quad sides\quad of\quad the\quad cube\quad at\quad the\quad corner,\quad the\quad distance\\ from\quad its\quad center\quad to\quad the\quad vertex\quad is\quad \sqrt { 3 } r.\quad \quad From\quad this,\quad we\\ can\quad add\quad up\quad everything\quad along\quad the\quad diagonal\quad of\quad the\quad cube\\ and\quad come\quad up\quad with\quad the\quad following\quad equation:\quad \\ \sqrt { 3 } =\sqrt { 3 } r+r+2r+r+\sqrt { 3 } r\\ Solving\quad for\quad r,\quad we\quad have\quad r=\frac { \sqrt { 3 } }{ 4+2\sqrt { 3 } }$

In $Oxyz$ coordinate system with $O$ is the origin $(0,0,0)$ , the center $O_1$ of the central sphere is $(\frac{1}{2},\frac{1}{2},\frac{1}{2})$ . One center (we call it $O_2$ ) of the other spheres must be $(t, t, t)$ and it's radius is also $t$ . We have $|O_1O_2|=2r=2t$ , that is

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

so $t$ must be the root of the above equation (which is a quadratic equation). We solve the equation and choose $t\in (0,\frac{1}{2})$

Taking a plane through the diagonal of the cube and a side. The section on the cube will be a rectangle with sides 1 and $\sqrt{2}$ . We will see three circles on it with radius say $r$ . The center of these circles will be on its diagonal with length $\sqrt{3}$ . The end circles will touch the $\sqrt{2}$ side, but will be away from the edge. this center will be at a distance of r from all three sides of the cube it touches. So it is at a distance of diagonal of a cube sides r. That is $\sqrt{3} * r$ . So the length of the diagonal of our cube will equal to distances of the edge circles from its vertexics plus the distance between the centers of these circles.

$\sqrt{3}=\sqrt{3}*r +4*r+ \sqrt{3}*r~~\implies r=0.232 \\ \boxed{0.23}\\$

This 2D method has the same logic as that of Michael Mendrin which is in 3D.

This problem is from 1990 AHSME.