3. Given a tetrahedron $S-ABC$ with the base being an isosceles right triangle with hypotenuse $AB$, and $SA=SB=SC=AB=2$. Find the surface area of the circumscribed sphere of the tetrahedron $S-ABC$ is $\qquad$ .
3. $\frac{16 \pi}{3}$.

As shown in Figure 3, let the circumcenter of the tetrahedron $S-A B C$ be $O$, and the radius be $R$.

Given $S A=S B=S C$ $=A B=2$, we know that the projection of $S$ on the plane $A B C$ is the circumcenter of $\triangle A B C$, which is the midpoint $H$ of $A B$.

Since $O A=O B=O C$, we know that the projection of $O$ on the plane $A B C$ is also the midpoint $H$ of $A B$.
Thus, $S, O, H$ are collinear.
Furthermore, since $O A=O B=O S$, we know that $O$ is the circumcenter of $\triangle S A B$.
Therefore, $R=O A=\frac{2}{3} \times \frac{\sqrt{3}}{2} \times 2=\frac{2 \sqrt{3}}{3}$.
Hence, the required result is $4 \pi \times\left(\frac{2 \sqrt{3}}{3}\right)^{2}=\frac{16 \pi}{3}$.