We know Borromean rings in a 3-sphere $S^3$ can be unlinked if we remove one of the three rings.
Here let us consider a slight different procedure. If we remove the neighbored solid torus $B^2 \times S^1$ region of one of the three rings from $S^3$ , then the remained space after removing $B^2 \times S^1$ out of $S^3$ is still another $S^1 \times B^2$, since the gluing along gives $(B^2 \times S^1) \cup (S^1 \times B^2)=S^3$.
Question: Is the remained two rings out for the original Borromean rings are linked or unlinked in the remained complement space $S^1 \times B^2$ (which is the remained space from $S^3-(B^2 \times S^1)$)?
Note: $B^2$ is a 2-ball or equivalently a 2-disk. Say we remove the blue ring within $S^1 \times B^2$ and the remained red and green rings are still in the remained complement space $S^1 \times B^2=S^3-(B^2 \times S^1)$.
As far as I understand the question, the answer is linked. You can't unlink the two remaining knots in the solid torus. An unlinking would give you an unlinking of the original problem, for which we know it does not exist.