Let $F$ denote this figure. Consider the translations of $F$ under the following vectors in the plane: $\vec{u}_1 = 0.001\vec{i}$, and
$$\vec{u}_{k+1} = R_{\pi/3}(\vec{u}_k), \quad k = 1, 2, 3, 4, 5,$$
where $R_{\pi/3}$ denotes the rotation by $\pi/3$ counterclockwise. Now let $F_0 = F$ and, for $k = 1, \dots, 6$,
$F_k$ = the translation of $F$ in the direction $\vec{u}_k$.

We will show that $F_k \cap F_l = \emptyset$ for all $k, l \in \{0, \dots, 6\}, k \neq l$. Since the distance between any two points of $F$ is not equal to $0.001$, $F_0$ and $F_k$ have no common points for $k = 1, \dots, 6$. By the same reasoning, $F_k$ and $F_{k+1}$ have no common points for $k = 1, \dots, 6$ ($F_7 = F_1$). Next since the distance between any two points in $F$ is not equal to $0.001 \times \sqrt{3}$, it follows that $F_k$ and $F_{k+2}$ have no common points. Finally, that the distance between any two points in $F$ is not equal to $0.002$ implies $F_k$ and $F_{k+2}$ have no points in common. Thus $F_k \cap F_l = \emptyset$ for all $k \neq l$ as claimed.

The figures $F_k$ are all lying in the square of side $5.002$, and they have pairwise empty intersection. Thus
$$7 \times \text{area}(F) = \sum_{i=0}^{6} \text{area}(F_i) \le 5.002^2;$$
therefore $\text{area}(F) \le 5.002^2/7 < 3.575$. $\square$