We have a rectangular block measuring L by W by H, and we know:

LW = 54
LH = 36
WH = 24

Notice if we multiply all three of LW, LH and WH together, we get

(LW)(LH)(WH) = (54)(36)(24)
(L^2 W^2 H^2) = (6)(9)(6^2)(6)(4)
(LWH)^2 = (2^2)(6^4)(3^2)
LWH = 2*6^2*3 = 6^3

So the volume of the rectangular block is 6^3. If we reshape it into a cube, the volume won't change (or at least, if it did, the question has no solution), so our cube has volume 6^3, and thus has edges of length 6, and a surface area of 6*6^2 = 6^3 = 216, since we have six square faces each with area 6^2.

There is an issue with the question, however. The only way to solve the question is to assume the volumes of the rectangular block and of the cube are the same. But when we melt iron, it expands -- the volume does change. If you use actual data about iron density here, the correct answer is more like 235, and not 216.

You can also pick what is likely the right answer without doing much work. The surface area of the original block is 2(24 + 36 + 54) = 228. When we reshape that into a cube, we'll get a lower surface area for the same volume. To see that, you can imagine instead flattening a rectangular block so the top and bottom faces become enormous and the height becomes minuscule -- then the surface area becomes huge for the same volume. Doing the reverse, i.e. making the edges all equal, will make the surface area smaller for the same volume. So we want an answer comparable in size to, but smaller than, 228, and only B or C are plausible. But if we divide each of B and C by 6, we get 26 and 36, and if we assume the edges of the resulting cube will turn out to be integers here (which would usually but not always be true on a GMAT question like this), C is the most likely answer.