The area of one triangular face of the pyramid is $\frac{1}{2}(s)(15)$. There are four congruent triangular faces.

After finding $s$, use a right triangle whose hypotenuse is the face altitude and whose horizontal leg is half the base side length. Then use the prism-volume and pyramid-volume formulas.

Algebra is quicker for this problem, but Desmos can verify the base side length. Enter $y=x^2+78x$ and $y=1728$. Select the intersection with positive $x$-coordinate; it represents the square's side length.

Enter sqrt(15^2-(18/2)^2) to verify the pyramid's vertical height using the Pythagorean theorem.

Enter 18^2*12+(1/3)*18^2*12 to calculate the sum of the prism's volume and the pyramid's volume.

Let $s$ be the side length of the square base. The exposed bottom has area $s^2$. The prism has four exposed lateral faces with total area $4(12s)=48s$. The pyramid has four triangular lateral faces with total area $4\left(\frac{1}{2}s\cdot15\right)=30s$.

Thus, $s^2+48s+30s=1728$, or $s^2+78s=1728$.

Rearrange and factor:

$s^2+78s-1728=0$

$(s-18)(s+96)=0$

A length must be positive, so $s=18$.

In a right square pyramid, the face altitude forms a right triangle with the pyramid's vertical height and half the base side length. Half of the base side is $\frac{18}{2}=9$.

Therefore, the vertical height $h$ satisfies $h^2+9^2=15^2$, so $h=12$.

The prism volume is $18^2(12)=3888$.

The pyramid volume is $\frac{1}{3}(18^2)(12)=1296$.

The total volume is $3888+1296=5184$.