Solution:

We distinguish 2 cases:

- if $a=2$ we must have $b^{c}<11$ because $2^{11}=2048>2002$, so $b<4$ and if $b=2$ then $c$ can be $2$ or $3$, while if $b=3$ then $c=2$: thus we obtain the solutions $(2,2,2)$, $(2,2,3)$, $(2,3,2)$;

- if $a>2$ we must have $b^{c}<7$ because $a^{7} \geq 3^{7}=2187>2002$. Therefore the only possibility is to have $b=2$ and $c=2$. Since $6^{4}=1296<2002$ while $7^{4}=2401>2002$, the admissible values are $a=3,4,5,6$ which give the solutions $(3,2,2)$, $(4,2,2)$, $(5,2,2)$, $(6,2,2)$ for a total of $7$ solutions.