smc

As the dimensions are in geometric progression, let them be $\frac{b}{r}$, $b$, and $br$ cm, so the volume is $\left(\frac{b}{r}\right)(b)(br)=b^3$, giving $b^3=8$ and thus $b=2$. The surface area condition now yields $$\begin{array}{rl}2\left(\frac{2}{r}\right)(2)+2(2)(2r)+2(2r)\left(\frac{2}{r}\right)=32& ⟺\frac{8}{r}+8+8r=32\\ & ⟺r+\frac{1}{r}=3\\ & ⟺r^2-3r+1=0\\ & ⟺r=\frac{3\pm \sqrt{5}}{2}.\end{array}$$ Since $$\frac{3-\sqrt{5}}{2}=\frac{1}{\left(\frac{3+\sqrt{5}}{2}\right)},$$ the two possible values of $r$ do not actually give different dimensions, but merely determine whether they are in increasing or decreasing order. Therefore, without loss of generality, we take $r=\frac{3+\sqrt{5}}{2}$, giving the dimensions (in cm) as $$\begin{array}{rl}& \frac{2}{\left(\frac{3+\sqrt{5}}{2}\right)},2,\text{ and }2\left(\frac{3+\sqrt{5}}{2}\right)\\ & =\frac{4}{3+\sqrt{5}},2,\text{ and }3+\sqrt{5}\\ & =3-\sqrt{5},2,\text{ and }3+\sqrt{5}.\end{array}$$ As there are $4$ edges with each of these distinct lengths, it follows that the sum of all the edge lengths (in cm) is $$\begin{array}{rl}4\left(3-\sqrt{5}+2+3+\sqrt{5}\right)& =4(8)\\ & =\boxed{32}.\end{array}$$