To find the slant height of one triangular wall, use the distance from the face to the cube's center and the distance from the center of the square opening to one of its sides.

After finding $x^2$, use the square-pyramid volume formula with height equal to half the cube's edge length.

Enter $y=2x\sqrt{81+\frac{x^2}{4}}-x^2\{0<x\le18\}$ and $y=108$. The positive intersection gives the possible side length of the square opening.

Use a Desmos calculation line to square the positive intersection's $x$-value. This verifies the value of $x^2$ needed for the pyramid-volume calculation.

Enter $18^3-3x^2$, using the valid intersection value for $x$. This subtracts the square pyramid's volume from the cube's volume.

The apex is $9$ centimeters from the face because it is at the cube's center. From the center of the square base to the midpoint of one side is $\frac{x}{2}$, so the slant height of each triangular interior wall is $\sqrt{9^2+\left(\frac{x}{2}\right)^2}$.

The square opening removes area $x^2$. The four triangular walls together have area $2x\sqrt{81+\frac{x^2}{4}}$. Therefore, $2x\sqrt{81+\frac{x^2}{4}}-x^2=108$.

Add $x^2$ to both sides and square: $4x^2\left(81+\frac{x^2}{4}\right)=(x^2+108)^2$. Simplifying gives $324x^2+x^4=x^4+216x^2+11664$, so $108x^2=11664$ and $x^2=108$.

The cavity is a square pyramid with height $9$, so its volume is $\frac{1}{3}x^2(9)=3(108)=324$. The cube's volume is $18^3=5832$, so the remaining volume is $5832-324=5508$.