4. As shown in the figure, in the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, it is known that $A B=B C=6, A A_{1}=2$, points $M, N$ are on edges $D A, D C$ respectively, and the dihedral angle $D_{1}-M N-D$ is $45^{\circ}$. If the volume of the tetrahedron $D_{1}-D M N$ is $\frac{16 \sqrt{3}}{9}$, then the surface area of the circumscribed sphere of the tetrahedron $D_{1}-D M N$ is
Draw $D H \perp M N$ at $H$, then $D H=$ $D D_{1}=2$, and $V_{D_{1}-D M N}=\frac{16 \sqrt{3}}{9}$ $\Rightarrow S_{\triangle D M N}=\frac{8 \sqrt{3}}{3} \Rightarrow M N=\frac{8 \sqrt{3}}{3}$,
Let $M H=s, H N=t$, then $s, t$ are the two roots of the equation $x^{2}-\frac{8 \sqrt{3}}{3} x+4=0$,
Solving, we get $M H=2 \sqrt{3}, H N=\frac{2 \sqrt{3}}{3} \Rightarrow D M=4, D N=\frac{4 \sqrt{3}}{3}$. Therefore, the surface area of the circumscribed sphere of the tetrahedron $D_{1}-$ $D M N$ is $S=\left(4+16+\frac{16}{3}\right) \pi=\frac{76}{3} \pi$.