Check the first picture below.
we know it's open at the top, thus we'll exclude the base atop from the surface area, we also know its volume is to be 4 cm³, and that its base is a square, so we can say that
[tex]\stackrel{\textit{volume of a rectangular prism}}{4 = x\cdot x\cdot h}\implies 4 = x^2h\implies \boxed{\cfrac{4}{x^2}=h} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{surface area of the prism}}{SA=(x\cdot x)+(x\cdot h)+(x\cdot h)+(x\cdot h)+(x\cdot h)} \\\\\\ SA = x^2 + 4xh\implies SA = x^2+4x\left( \cfrac{4}{x^2} \right)\implies SA(x) = x^2+16x^{-1} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\stackrel{\textit{getting its derivative}}{\cfrac{dSA}{dx}=2x-\cfrac{16}{x^2}}\implies \cfrac{dSA}{dx}=2\left( x-\cfrac{8}{x^2} \right) \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{setting its derivative to 0}}{0=2\left( x-\cfrac{8}{x^2} \right)}\implies \cfrac{8}{x^2}=x\implies 8 = x^3\implies \sqrt[3]{8}=x\implies 2 = x[/tex]
now if we do a first derivative test on that critical point of 2, hmmm the way I usually do it is, using a calculator, subtract a tiny bit from it, give it to a variable, to check the region on the left, and then add a tiny bit to it, give it to a variable, to check the region to its right.
so if we do that, say hmmmm 2 - 0.0000001 to some variable and then 2 + 0.0000001 to some other variable, plug those variables in the first derivative, then we'll see that, the region of the left of it is negative, and to its right is positive, check the 2nd picture below, meaning is a minimum at the critical point of 2, and its area will be
[tex]SA(2)=2^2+\cfrac{16}{2}\implies SA(2) = 4+8\implies \boxed{SA = 12}[/tex]
