Answer: 288

Small Hint:

Find the height by projecting either endpoint of the upper edge onto the base

Big Hint:

At a fraction $t$ of the height, the cross-section is a rectangle with sides $s(1-t)$ and $s(1+t)$

Solution:

Each endpoint of the upper edge is joined to the endpoints of one side of the square base. Its distance to that side’s midpoint is therefore $\frac{\sqrt3s}{2}.$ The projection of each upper endpoint lies $\frac{s}{2}$ beyond that midpoint, because the two projected endpoints are $2s$ apart while the two opposite-side midpoints are $s$ apart. Thus the height $h$ satisfies $$h^2=\left(\frac{\sqrt3s}{2}\right)^2-\left(\frac{s}{2}\right)^2 =\frac{s^2}{2},$$ so $h=\frac{s}{\sqrt2}.$

At a fraction $t$ of the height above the base, the cross-section is a rectangle whose dimensions are $s(1-t)$ and $s(1+t).$ Its area is $s^2(1-t^2).$ By Cavalieri’s principle, the volume is the volume $s^2h$ of a prism minus the volume $\frac{s^2h}{3}$ of a pyramid: $$V=\frac23s^2h=\frac{\sqrt2}{3}s^3.$$ Substituting $s=6\sqrt2$ gives $V=288.$