A cube with edge 2 cm has spheres, each of radius 1 cm, at each vertex...
Let's consider one of the spheres at the vertex of the cube. We draw a line connecting the center of this sphere to the center of the cube.
[asy]
size(100);
import three;
triple A=(0,0,0),B=(0,1,0),C=(1,1,0),D=(1,0,0),E=(0,0,1),F=(0,1,1),G=(1,1,1),H=(1,0,1);
triple P=(0.5,0.5,0),Q=(0.5,0.5,1);
draw(P--Q,dashed);
draw(A--B--C--D--A--E--F--G--H--E--F--B--C--G--H--D,dashed);
draw(A--E--P--B--F--Q--C--G--D--H--cycle);
draw(Q--H--G--C--Q);
draw(P--A,dashed);
draw(P--B,dashed);
draw(P--C,dashed);
draw(P--D,dashed);
draw(Q--E,dashed);
draw(Q--F,dashed);
draw(Q--G,dashed);
draw(Q--H,dashed);
dot((0,0,0));
dot((1,0,0));
dot((0,1,0));
dot((1,1,0));
dot((0,0,1));
dot((1,0,1));
dot((0,1,1));
dot((1,1,1));
dot((0.5,0.5,0));
dot((0.5,0.5,1));
[/asy]
Let's call the center of this sphere $P$ and the center of the cube $Q$. We want to find the radius of the largest sphere that can be centered within the cube and touches each of the remaining spheres externally in at most one point.
To do this, we need to find the distance between $P$ and $Q$.
We can see that $PQ$ is a diagonal of the cube, so we can use the Pythagorean theorem to find its length.
The length of one side of the cube is $2$ cm, so the length of $PQ$ is $\sqrt{2^2 + 2^2 + 2^2} = \sqrt{12} = 2\sqrt{3}$ cm.
Therefore, the radius of the largest sphere that can be centered within the cube and touches each of the remaining spheres externally in at most one point is $\boxed{\sqrt{3}}$ cm.