3. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with hypotenuse $AB$, and $SA=SB=SC=AB=2$. Suppose points $S, A, B, C$ all lie on a sphere with center $O$. Then the surface area of this sphere is $\qquad$
3. $\frac{16 \pi}{3}$.

Since $S A=S B=S C$, the projection of point $S$ on the base, $D$, is the circumcenter of $\triangle A B C$, which is the midpoint of $A B$. It is easy to see that point $O$ lies on $S D$ and is the centroid of $\triangle S A B$. We can find that $O S=\frac{2 \sqrt{3}}{3}$. Therefore, the surface area of sphere $O$ is $4 \pi R^{2}=4 \pi\left(\frac{2 \sqrt{3}}{3}\right)^{2}=\frac{16 \pi}{3}$.