Solution:

Since $216 = 2^{3} 3^{3}$, any positive divisor of $216$ must be of the form $2^{x} 3^{y}$ for some integers $x$ and $y$ with $0 \leq x, y \leq 3$. Thus, we set $a = 2^{x_{1}} 3^{y_{1}}$, $b = 2^{x_{2}} 3^{y_{2}}$, $c = 2^{x_{3}} 3^{y_{3}}$ and $d = 2^{x_{4}} 3^{y_{4}}$, where $0 \leq x_{i}, y_{i} \leq 3$ are integers for $i = 1, \ldots, 4$. We compute
$$2^{3} 3^{3} = 216 = a b c d = 2^{x_{1} + x_{2} + x_{3} + x_{4}} 3^{y_{1} + y_{2} + y_{3} + y_{4}}$$
so the number of such ordered quadruples $(a, b, c, d)$ is the number of ordered 8-tuples $(x_{1}, x_{2}, x_{3}, x_{4}, y_{1}, y_{2}, y_{3}, y_{4})$ of nonnegative integers such that $x_{1} + x_{2} + x_{3} + x_{4} = y_{1} + y_{2} + y_{3} + y_{4} = 3$. By stars-and-bars, this number is $\binom{3 + 4 - 1}{4 - 1}^{2} = 20^{2} = 400$.