@Sp3000 is right, this is actually $PE-163$, and your particular case is given in the problem statement $ T(2) = 104.$
But if you are looking for a general formula to count the number of triangles in higher order then check here, (spoiler for the original PE problem).
Say that instead of four triangles along each edge we have $n$. First count the triangles that point up. This is easy to do if you count them by top vertex. Each vertex in the picture is the top of one triangle for every horizontal grid line below it. Thus, the topmost vertex, which has $n$ horizontal gridlines below it, is the top vertex of $n$ triangles; each of the two vertices in the next row down is the top vertex of $n-1$ triangles; and so on. This gives us a total of
$$\begin{align*}
\sum_{k=1}^nk(n+1-k)&=\frac12n(n+1)^2-\sum_{k=1}^nk^2\\
&=\frac12n(n+1)^2-\frac16n(n+1)(2n+1)\\
&=\frac16n(n+1)\Big(3(n+1)-(2n+1)\Big)\\
&=\frac16n(n+1)(n+2)\\
&=\binom{n+2}3
\end{align*}$$
upward-pointing triangles.
The downward-pointing triangles can be counted by their by their bottom vertices, but it’s a bit messier. First, each vertex not on the left or right edge of the figure is the bottom vertex of a triangle of height $1$, and there are $$\sum_{k=1}^{n-1}=\binom{n}2$$ of them. Each vertex that is not on the left or right edge or on the slant grid lines adjacent to those edges is the bottom vertex of a triangle of height $2$, and there are
$$\sum_{k=1}^{n-3}k=\binom{n-2}2$$ of them. In general each vertex that is not on the left or right edge or on one of the $h-1$ slant grid lines nearest each of those edges is the bottom vertex of a triangle of height $h$, and there are
$$\sum_{k=1}^{n+1-2h}k=\binom{n+2-2h}2$$ of them.
Algebra beyond this point corrected.
The total number of downward-pointing triangles is therefore
$$\begin{align*}
\sum_{h\ge 1}\binom{n+2-2h}2&=\sum_{k=0}^{\lfloor n/2\rfloor-1}\binom{n-2k}2\\
&=\frac12\sum_{k=0}^{\lfloor n/2\rfloor-1}(n-2k)(n-2k-1)\\
&=\frac12\sum_{k=0}^{\lfloor n/2\rfloor-1}\left(n^2-4kn+4k^2-n+2k\right)\\
&=\left\lfloor\frac{n}2\right\rfloor\binom{n}2+2\sum_{k=0}^{\lfloor n/2\rfloor-1}k^2-(2n-1)\sum_{k=0}^{\lfloor n/2\rfloor-1}k\\
&=\left\lfloor\frac{n}2\right\rfloor\binom{n}2+\frac13\left\lfloor\frac{n}2\right\rfloor\left(\left\lfloor\frac{n}2\right\rfloor-1\right)\left(2\left\lfloor\frac{n}2\right\rfloor-1\right)\\
&\qquad\qquad-\frac12(2n-1)\left\lfloor\frac{n}2\right\rfloor\left(\left\lfloor\frac{n}2\right\rfloor-1\right)\;.
\end{align*}$$
Set $\displaystyle m=\left\lfloor\frac{n}2\right\rfloor$, and this becomes
$$\begin{align*}
&m\binom{n}2+\frac13m(m-1)(2m-1)-\frac12(2n-1)m(m-1)\\
&\qquad\qquad=m\binom{n}2+m(m-1)\left(\frac{2m-1}3-n+\frac12\right)\;.
\end{align*}$$
This simplifies to $$\frac1{24}n(n+2)(2n-1)$$ for even $n$ and to
$$\frac1{24}\left(n^2-1\right)(2n+3)$$ for odd $n$.
The final figure, then is
$$\binom{n+2}3+\begin{cases}
\frac1{24}n(n+2)(2n-1),&\text{if }n\text{ is even}\\\\
\frac1{24}\left(n^2-1\right)(2n+3),&\text{if }n\text{ is odd}\;.
\end{cases}$$
Best Answer
Let's see! We'll break it up into the following cases (where an apex means one of the six points of the star):
A: Triangles that contain three apexes
There are 2 of these.
B: Triangles that contain a pair of opposite apexes
For each such pair, there are 2 of these, so 6 in total.
C: Triangles that contain a pair of non-opposite apexes
For each non-adjacent apex pair, there are three of these, so 18 in total.
D: Triangles that contain exactly one apex
For each apex, I count 10 such triangles, so 60 in total.
E: Triangles that contain no apex
For each edge of the internal hexagon, there are 3 triangles, so 18 in total.
I make that 2 + 6 + 18 + 60 + 18 = 104.