As shown in Figure 1, take the midpoint $M$ of $BC$, connect $AM$ and extend it to $D$ such that $AD=4$. It is easy to prove that $\triangle ABD$ and $\triangle ACD$ are both right triangles. Therefore, quadrilateral $ABDC$ can be inscribed in a circle. Thus, we can use quadrilateral $ABDC$ as the base to complete the right prism $ABC-A_{1}B_{1}C_{1}$ into a right prism $ABDC-A_{1}B_{1}D_{1}C_{1}$ inscribed in a sphere. From the given conditions, it is easy to find that the body diagonal $A_{1}D=2\sqrt{5}$. Therefore, the radius of the circumscribed sphere is $R=\sqrt{5}$, so the surface area of the sphere $S=4\pi R^{2}=20\pi$. The answer should be $20\pi$.