Solution:

The answer is $29$. The figure drawn on the sheet can be represented as follows:

The point $O$ in the figure is given by the center of the larger hexagon $A B C D E F$. Since the hexagon $A B C D E F$ is regular, the triangle $O C D$ is equilateral. Moreover, since $D A$ is the diameter of the circle circumscribed to $A B C D E F$, the triangle $A C D$ is right-angled, with the right angle at $C$ and $A D C=60^{\circ}$. In particular, denoting by $\ell$ the length of the side $D C$ of the hexagon, we have $A D=2 \ell$. By the Pythagorean theorem, we have $A C=\sqrt{3} \ell$; it follows that the ratio between the major diagonal of the smaller hexagon $A K G C J I$ and the major diagonal of the hexagon $A B C D E F$ is $\frac{\sqrt{3}}{2}$ and, therefore, the ratio of the respective areas is $\frac{3}{4}$, from which
$$\operatorname{Area}(A B C D E F)=\frac{4}{3} \operatorname{Area}(A K G C J I)=\frac{4}{3} \times 18=24$$
The trapezoid $A I J C$ is half of the hexagon $A K G C J I$, from which
$$\operatorname{Area}(A I J C)=\frac{1}{2} \operatorname{Area}(A K G C J I)=\frac{1}{2} \times 18=9.$$
The quadrilateral $O C B A$ is one third of the hexagon $A B C D E F$ (indeed it is the union of the two equilateral triangles $O A B$ and $O B C$, both corresponding to $\frac{1}{6}$ of the hexagon); we obtain
$$\operatorname{Area}(O C B A)=\frac{1}{3} \operatorname{Area}(A B C D E F)=\frac{1}{3} \times 24=8$$
Note that the triangles $O A C$ and $B A C$ are congruent: indeed $\angle C B A=120^{\circ}=\angle C O A$, they have $C A$ as a common side and are both isosceles; from which
$$\operatorname{Area}(A B C)=\frac{1}{2} \operatorname{Area}(O C B A)=\frac{1}{2} \times 8=4.$$
It follows directly that the area of the pentagon $A C D E F$ is given by
$$\operatorname{Area}(A C D E F)=\operatorname{Area}(A B C D E F)-\operatorname{Area}(A B C)=24-4=20.$$
Finally, the area of the union of the two hexagons is given by the sum of the areas of the pentagon $A C D E F$ and of the trapezoid $A I J C$, that is
$$\operatorname{Area}(A I J C D E F)=\operatorname{Area}(A I J C)+\operatorname{Area}(A C D E F)=9+20=29.$$