Answer     $3$

The given relation can be rewritten as $$N^{\frac{1}{a}+\frac{1}{ab}+{1}{abc}}=N^{\frac{25}{36}}\implies \frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{36}\implies\frac{bc +c + 1}{abc}=\frac{25}{36}$$

Assuming $$abc=36\quad\text{and}\quad bc+c+1=25$$

Then we have $$bc+c=24\implies (b+1)c=24\implies \frac{abc}{(b+1)c}=\frac{36}{24}\Leftrightarrow \frac{ab}{b+1}=\frac{3}{2}$$

Also, by the assumption $abc=36$, we know $a$, $b$, $c$ are some combination of $2$ and $3$. There are a limit number of possibilities. Try them out will lead to the solution $(a, b, c) = (2, 3, 6)$.