Solution:

The answer is (B). For $n \geq 1$ the figure $F_{n}$ is a hexagon having three long sides and three short sides, alternating with each other. Each of the long ones is made up of $n+1$ sides of triangles, while each of the short ones is made up of $n$ sides of triangles. Letting $x_{n}$ be the number of triangles of the tessellation making up $F_{n}$, we compute the analogous number $x_{n+1}$ by counting the triangles of the frame that is added to $F_{n}$ to obtain $F_{n+1}$. This count gives $x_{n+1}-x_{n}$. We can build the frame in three stages: in the first we add one triangle for each of the sides of the triangles that make up the sides of the hexagon $F_{n}$, thus altogether $3(2 n+1)$ triangles; in the second stage we add the triangles that must be inserted between those of the first stage, $n$ on each long side and $n-1$ on each short side, thus altogether $3(2 n-1)$; finally, in the third stage we add two triangles for each vertex of $F_{n}$, that is 12 triangles. The total is
$$x_{n+1}-x_{n}=3(2 n+1)+3(2 n-1)+12=12 n+12 \quad \text{cioè} \quad x_{n+1}-x_{n}=12(n+1)\text{.}$$
Let us now write $x_{1}=1+12$ (the result of an easy direct count) and the previous formula for $n=1, \ldots, 9$ (that is $x_{2}-x_{1}=12 \cdot 2, \ldots x_{10}-x_{9}=12 \cdot 10$ ), sum everything term by term and simplify the left-hand side. What remains is that $x_{10}=1+12+12 \cdot 2+12 \cdot 3+\cdots+12 \cdot 10$. Hence
$$x_{10}=1+12(1+2+3+\cdots+10)=1+12 \cdot \frac{10 \cdot 11}{2}=661 .$$

Second Solution

The figure $F_{n}$ is an equilateral triangle of side $3 n+1$ from which 3 equilateral triangles of side $n$ resting on the vertices have been removed. Hence $F_{n}$ is made up of $(3 n+1)^{2}-3 n^{2}=6 n^{2}+6 n+1$ small triangles of side 1 .