jmc

First, take a cross-section through the plane of the hexagon: Since the side length of the hexagon is $2,$ the distance from the center of the hexagon to each of its vertices is also $2.$ Combined with the fact that the radius of each of the smaller spheres is $1,$ we see that the radius of the larger sphere is $2+1=3.$

To find the radius of the eighth sphere, we take a cross-section perpendicular to the plane of the hexagon, which passes through two opposite vertices of the hexagon: (Here $A$ is a vertex of the hexagon, $O$ is the center of the hexagon, and $P$ is the center of the eighth sphere.) Let $r$ be the radius of the eighth sphere, centered at $P.$ Then $AP=r+1$ and $OP=3-r$ (since the radius of the larger sphere is $3$). We also know that $AO=2,$ since $A$ is a vertex of the hexagon and $O$ is the center of the hexagon. Thus, by Pythagoras, $$2^2+(3-r{)}^2=(r+1{)}^2,$$or $r^2-6r+13=r^2+2r+1.$ Then $8r=12,$ and so $r=\boxed{\frac{3}{2}}.$