The top of the prism and the base of the pyramid touch each other inside the solid, so neither is part of the exterior surface area.

After finding $s$, use the pyramid's vertical height and half the base side length to form a right triangle for the slant height.

Algebra is quick for this problem. For a Desmos verification, graph $y=\frac{9}{4}x^3$ and $y=1152$. Use the positive intersection to find the value of $s$.

In a new expression line, enter sqrt(6^2+4^2) to verify that the slant height of a triangular pyramid face is $2\sqrt{13}$ after using the value of $s$.

Enter 8*8+4*8*16+4*(1/2)*8*2sqrt(13) to add the bottom of the prism, its four lateral faces, and the pyramid's four triangular faces.

The prism has volume $s^2(2s)=2s^3$.

The pyramid has volume

Therefore,

So $\frac{9}{4}s^3=1,152$, which gives $s^3=512$ and $s=8$.

The prism height is $2s=16$, and the pyramid's vertical height is $\frac{3}{4}s=6$.

For one triangular face of the pyramid, the horizontal distance from the center of the square base to the midpoint of a side is half of $8$, or $4$. Thus, the slant height is

The exposed bottom of the prism has area $8\cdot 8=64$.

The four lateral faces of the prism have total area $4(8\cdot16)=512$.

Each triangular face of the pyramid has area

so the four triangular faces have total area $32\sqrt{13}$.

Thus, the exterior surface area is