Consider the sphere: Compute $r$ from $V$ . We have

$288\pi =\dfrac{4}{3} \pi (r^3)$ $\implies$ $r=6$

So the surface area of the sphere is $S_{sphere}=4 \pi r^2=4(\pi)(6^2)\approx 452.39$

Consider the cube: Compute $a$ from $V$ , we have

$288 \pi=a^3$ $\implies$ $a\approx 9.67$

The surface area of the cube $S_{cube}=6(9.67^2) \approx 561.05$

$\color{#D61F06}\large{\boxed{\therefore \text{The cube has a larger surface area.}}}$

Firstly, we are given that the Sphere is melted down and reshaped into a Cube, with no waste. Therefore their volumes must be the same. So we have (using the equations given in the question): $\frac { 4 }{ 3 } \pi { r }^{ 3 }=288\pi \\ { a }^{ 3 }=288\pi$ Therefore we can say that the radius of the sphere and the side-length of the cube is: $r=6\\ { a }=9.67$ Therefore, using the formulas given, we can calculate the surface areas to be: ${ A }_{ sphere }=4\pi r^{ 2 }=4\cdot 3.14\cdot 6^{ 2 }=452cm^{ 3 }\\{ A }_{ cube }=6a^{ 2 }=6\cdot 9.67^{ 2 }=561cm^{ 3 }$ Therefore the cube has a greater surface area than the sphere, so the correct answer to the question is: $\boxed {The \ Cube}$

Obviously, the cube has more surface area as spheres have the least amount of surface area to pack a certain volume, as we see in soap bubbles. (We can't have a bubble with a shape other than sphere)

volume of sphere = 288π

4/3 * π * r^3 = 288π

r = 6 units

surface area of a sphere = 4 * π * r^2 = 4 * π * 6^2 = 452.39 square units

volume of cube = volume of sphere = 288π

s^3 = 288π

s = 9.67 units

surface area of cube = 6 * s^2 = 6 * 9.67^2 = 561.05 square units

surface area of cube > surface area of sphere

It's significant to consider the Area - Volume ratios of the sphere and cube:

A/V ratio of the sphere is:

$\frac{Asf}{Vsf}=\frac{4\pi r^{^{2}}}{\frac{4}{3}\pi r^{3}} = \frac{3}{r}$ ;

while

A/V ratio of the cube is:

$\frac{Ac}{Vc}=\frac{6l^{2}}{l^{3}}=\frac{6}{l}\$ Equation (1).

Volumes are equal so

$\frac{4}{3}\pi r^{3}=l^{3}\Rightarrow l=r\left (\frac{4}{3}\pi \right )^{\frac{1}{3}}$ Equation (2) .

So the A/V ratio of the cube EQ (1) can be written as a function of r using Equation 2:

$\frac{Ac}{Vc}={\frac{6}{\left (\frac{4}{3}\pi \right )^{\frac{1}{3}}r}}\approx \frac{3.72}{r}$ .

So if Vc = Vsf then Ac = 1.24 Asf, approximately .

Mathematically, the cube has a higher surface area by the equations in other answers. However, the problem is flawed. There had to have been chocolate that accidentally didn't make it into the cube - it's a law of cooking with chocolate, some ends up in your stomach.