$2$

$ab - c = 3$ $\implies$ $ab=3+c$ $abc = 18$ $\implies$ $ab=\frac{18}{c}$ $ab=ab$ $3+c=\frac{18}{c}$ $c^2+3c-18=0$ $(c+6)(c-3)=0$ $\implies$ $c+6=0$ $\wedge$ $c-3=0$ $c=-6$ $\wedge$ $c=3$ Since $c\in$$R^+$ we get that $c=3$ $ab=\frac{18}{c}$ $ab=6$ $\frac{ab}{c}=2$