1. We start with the given equations:
$$ab - c = 3$$
$$abc = 18$$

2. From the first equation, we can express $ab$ in terms of $c$:
$$ab = 3 + c$$

3. Substitute $ab = 3 + c$ into the second equation:
$$abc = 18$$
$$c(3 + c) = 18$$

4. Simplify the equation:
$$3c + c^2 = 18$$

5. Rearrange the equation to form a standard quadratic equation:
$$c^2 + 3c - 18 = 0$$

6. Solve the quadratic equation using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 3$, and $c = -18$:
$$c = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-18)}}{2 \cdot 1}$$
$$c = \frac{-3 \pm \sqrt{9 + 72}}{2}$$
$$c = \frac{-3 \pm \sqrt{81}}{2}$$
$$c = \frac{-3 \pm 9}{2}$$

7. This gives us two potential solutions for $c$:
$$c = \frac{6}{2} = 3$$
$$c = \frac{-12}{2} = -6$$

8. Since $c$ must be a positive real number, we discard $c = -6$ and keep $c = 3$.

9. Substitute $c = 3$ back into the expression for $ab$:
$$ab = 3 + c = 3 + 3 = 6$$

10. Finally, calculate $\frac{ab}{c}$:
$$\frac{ab}{c} = \frac{6}{3} = 2$$

The final answer is $\boxed{2}$.