Solution:

The answer is 1536.

Let $A B C D E F$ be a regular hexagon representing the table; the tablecloth of area $1024~\mathrm{cm}^2$ covers exactly the rectangle $A B D E$. Letting $O$ be the center of the hexagon and $H, K$ be the points of intersection between the segments $B D$ and $O C$, $A E$ and $O F$, respectively, the triangles $O A K$, $O E K$, $O B H$ and $O D H$ are all congruent and have an area equal to half the area of the equilateral triangles $A B O$ and $D E O$. It follows that the area of the rectangle $A B D E$ equals four times the area of $A B O$, while the area of the table equals six times the area of $A B O$. We conclude that the area of the table is $\frac{6}{4} \cdot 1024~\mathrm{cm}^2 = 1536~\mathrm{cm}^2$.