Suppose that the dimensions of the open container are $x$, $y$ and $z$ (in meters, I'll ignore units henceforth) with $z$ being the height. Then first, as a simple exercise, convince yourself that the volume of the container is $V = xyz$ and the surface area $S = xy + 2(yz+zx)$.
To see why the surface area formula is true, imagine painting the container from the outside, and list the faces that you would need to paint. In particular, how many faces would you need to paint? (Remember that the container is open.)
Going back to the problem, one can rewrite the given volume condition to be $z = \frac{V}{xy} = \frac{32}{xy}$. Plugging this in the formula for the surface area, we get $S(x,y) = xy + 2(y \cdot \frac{32}{xy}+ x \cdot \frac{32}{xy}) = \ldots$ (I'll skip simplifying this further).
Thus the given problem reduces to finding the minimum value of $S(x,y)$ subject to the constraints $x > 0$ and $y > 0$. (Note that the volume condition is not explicitly relevant to us anymore.) Can you take it from here?
Note. The formula for $S$ I have written down considers only the external surface. But since the container is open, it might be more sensible to speak of the total, i.e., external+internal, surface area. In this case, one would need to multiply the formula for $S$ by $2$. (How would this affect the answers?)