2013 AMC 10A Problem 22

The centers of the six radius-$1$ spheres form a regular hexagon of side length $2$, so each is $2$ units from the large sphere’s center. Hence the large sphere has radius $3$. Let the eighth sphere have radius $r$, and let its center be distance $x$ from the large sphere’s center. Internal tangency gives $x+r=3$, so $x=3-r$. Using the right triangle between the large center, a small-sphere center, and the eighth-sphere center, $(r+1)^2=2^2+(3-r)^2$. Thus $2r+1=13-6r$, so $r=\frac32$. Thus, B is the correct answer.