Solution:

The answer is $27$. Let us call $ABCD$ the base of the pyramid and $V$ its vertex far from the ground. Note that the shadow $V'$ of the point $V$ falls on a circle of radius $15~\mathrm{m}$ centered at the center of $ABCD$, and depending on the orientation of south with respect to the pyramid (which we consider fixed) it can fall on any point of the circle whatsoever. The shadow of the pyramid (this time including the ground area beneath the pyramid itself) is formed by the union of the four triangles $ABV'$, $BCV'$, $CDV'$, $DAV'$.

This shadow can take two remarkably different shapes. Refer to the figures: when $V'$ falls in one of the dashed parts of the circle in the top figure, the shadow is a pentagon (left figure); when $V'$ instead falls in one of the other parts, the shadow is a quadrilateral (right figure). Let us call $W$ (resp. $X$) the intersection of the ray $DA$ (resp. $BA$) with the circle. We want to show that in both configurations the area of the shadow is greater than or equal to the area of the trapezoid $WBCD$. Indeed, in the configuration of the left drawing, $S(\text{ombra}) = S(ABCD) + S(ABV')$, and $S(ABV') \geq S(ABW)$ because these are two triangles with equal base $AB$, and the height of $ABV'$ is greater than that of $ABW$. Similarly, in the configuration of the left drawing, $S(\text{ombra}) = S(BCD) + S(BDV')$, and $S(BDV') \geq S(BDW)$ because these are two triangles with equal base $BD$, and the height of $BDV'$ is greater than that of $BDW$ (to see this it suffices to note that $WX$ is parallel to $BD$). Hence the shadow of smaller area occurs when $V' = W$, or in the symmetric configurations on the other sides. We must therefore calculate the area of triangle $ABW$. Letting $M$ be the midpoint of $AD$, we have that $MW \perp OM$, $OW = 15~\mathrm{m}$, $OM = 9~\mathrm{m}$, from which by the Pythagorean theorem $MW = \sqrt{OW^2 - OM^2} = 12~\mathrm{m}$. Hence $AW = MW - AM = 12 - 9 = 3~\mathrm{m}$, and $S(ABW) = \frac{AB \cdot AW}{2} = \frac{18 \cdot 3}{2} = 27~\mathrm{m}^2$.