Solution:

The answer is (A). Let us denote by $A, B, C, D$ the vertices of the trapezoid, by $O$ the point of intersection of the diagonals, and by $H$ and $K$ the feet of the perpendiculars to $AC$ drawn, respectively, from $B$ and $D$. The area $S$ of the trapezoid $ABCD$ equals the sum of the areas of the triangles $ACB$ and $ACD$, which have the same base $AC$ and heights $BH$ and $CK$ respectively. Let us observe further that, since $A\widehat{O}B = D\widehat{O}C = 30^{\circ}$, $BH = BO / 2$ and $DK = OD / 2$, hence $BH + DK = (BO + OD) / 2 = 3~\mathrm{cm}$. Therefore
$$S = \frac{AC \cdot BH + AC \cdot DK}{2} = \frac{AC \cdot (BH + DK)}{2} = \frac{6~\mathrm{cm} \cdot 3~\mathrm{cm}}{2} = 9~\mathrm{cm}^{2}.$$