There are two types of triangles; those with the point up, and those with the point down. Since everything is symmetric, so all we need to do is count triangles pointing up (or down) and double it. We will count the ones pointing up.
Next, see that the six pointed star is just the superposition of one large triangle pointing up over one large triangle pointing down.
Now, we can start counting the up triangles.
Individual single-unit triangles pointing up are found in the large triangle pointing up, plus $3$ left over around the edge. The large triangle pointing up has a base of $6$. So, the number of individual triangles is $T_1=6+5+4+3+2+1+3=21+3=24$.
The next size of triangles are ones with a base of 2. These triangles contain 4 triangles, with an upside down triangle at the centre. Looking at the figure, we therefore only need to count the interior upside down triangles. This time, there are no triangles around the edge, and less of them can fit in the large triangle. We get $T_2=5+4+3+2+1=15$.
Repeat with the rest, and you get $T_3=4+3+2+1=10$, $T_4=3+2+1=6$, $T_5=2+1=3$ and $T_6=1$.
The sum of these is $T=24+15+10+6+3+1=59$. Double this to get the total and you have $118$.