jmc

The amount of steel used to create one ball with radius 1 is $\frac{4}{3}\pi (1^3)=\frac{4}{3}\pi$; the amount of steel used to create eight of these balls is $8\cdot \frac{4}{3}\pi =\frac{32}{3}\pi$.

Let the radius of the large steel be $r$. We have $\frac{4}{3}\pi r^3=\frac{32}{3}\pi$; solving for $r$ yields $r^3=8\Rightarrow r=2$. Thus the radius of the large ball is $\boxed{2}$ inches.