I think what you mean is that you have a cube of $\textbf{volume}$ $ \ 8\text{cm}^3$=2cm x 2cm x 2cm.
A cube has 6 faces of equal area.
Therefore, the total surface area of the cube is equal to 6 multiplied by the area of one of the faces.
The area of one of the faces is 4$\text{cm}^2$ = 2cm x 2cm.
Therefore the total area of the cube is 24$\text{cm}^2$ = 6 x 4$\text{cm}^2$.
One thing you must remember is your units of measurement, writing 8cm2 (2x2x2) really makes no sense, since you are equating an area with a single number without units.
The result of your internet search has returned the method for working out the volume of the cube (or in fact a cuboid).
Volume = Length x Width x Height $ \ \ $
(cm$^3$ = cm x cm x cm)
Although you derivation takes on account both the Surface area and the Volume, I'd like to put this solution in prospective of the second term.
Since I do not know the value of the sides $\ell_{\mathrm{box}}$ an the volume $\ell_{\mathrm{box}}^3$, but I know his weight, then the only factor that correlates them is the Density, which I'll call $\varrho_x$, expressed by:
$$
\varrho_x=\frac{m_{\mathrm{box}}}{\ell_{\mathrm{box}}^3}
$$
Now, if the metal is the same, and so $\varrho_x$ doesn't change, then I could write a simple equation:
$$
\varrho_x=\frac{m_{\mathrm{box}(i)}}{\ell_{\mathrm{box}(i)}^3}=\frac{m_{\mathrm{box}(f)}}{(2\ell_{\mathrm{box}(i)})^3}
$$
The only unknown is $m_{\mathrm{box}(f)}$, so:
$$
m_{\mathrm{box}(f)}=\frac{m_{\mathrm{box}(i)}\cdot 8\ell_{\mathrm{box}}^3}{\ell_{\mathrm{box}(i)}^3}=m_{\mathrm{box}(i)}\cdot 8
$$
As you can see, the difference between the surface area is related to the difference of the volume, so you have to multiply by 8, as this simple $2^3$ times the side of the original lenght.
Let's take a generalized example.
If you imagine one simple cube, with lenght $a$, then the volume and the surface area are:
$$
S=6\cdot a^2,\quad V=a^3
$$
It's true that they have different factors, but they do not mean the same thing: if you imagine the increasing of weight, it's only given by the infinite variation of the Volume, as a simple integral:
$$
V_f=V_i+\left(\int_{a}^{2a}\mathrm{d}a\right)^3
$$
Above it can be observed that one single infintesimal change into the third dimension it's different from a single change of the surface; because the expansion is also inside the cube, which is not counted in the "$4$" factor, as two dimensional.
As you measure only the external surface, the volume takes the "internal expansion" into account, with either the relation of the density and the factor of $2$ "cubed" into the third dimension.
Best Answer
What you have is volume as a function of surface area. You want the other way around, meaning you'll need to invert what you have: $$ \textsf{SurfaceArea} = 6(\textsf{Volume})^{2/3} $$