I approached this way:
Having equal sides $1,2,3,4$ then the possible $3$ digit numbers are $111,112,221,222,223,...,448$
Then total number of $3$ digit numbers possible
$2(1+2+3+4)=20$
$20-4=16$
$16\cdot3=48$
Now for equal sides $5,6,7,8,9$
$5\cdot9=45$ ,$45-5=40,40\cdot3=120$
$120+48=168$
But the solution given is $165$.Even after checking a lot I couldn't get where is wrong.
I can only find $1+3+5+7 = 16$ triangles with equal sides $a$ and $b$ of lengths $1,\,2,\,3,\,4$ which respect the triangle inequality. From those $16$ triangles, I subtract the four equilateral ones, permute the remaining ones and add the equilateral triangles again, which gives $$ (16-4)\cdot 3 + 4 = 40 $$ triangles with equal sides $1,\,2,\,3,\,4.$
There are $45$ triangles with equal sides $a$ and $b$ of lengths $5,\,6,\,7,\,8,\,9$ which respect the triangle inequality. From those $45$ triangles, I subtract the five equilateral ones, permute the remaining ones and add the equilateral triangles again, which gives $$ (45-5)\cdot 3 + 5 = 125 $$ triangles with equal sides $5,\,6,\,7,\,8,\,9.$