Explanation:

- First, calculate the volume of the cuboid:

\( 36 \, \text{cm} \times 54 \, \text{cm} \times 24 \, \text{cm} = 46656 \, \text{cm}^3 \).

- Since the cuboid is melted and recast into 8 cubes of the same volume, each cube has a volume of:

\( 46656 \, \text{cm}^3 \div 8 = 5832 \, \text{cm}^3 \).

- The side length of each cube is the cube root of its volume:

\( \sqrt[3]{5832} = 18 \, \text{cm} \).

- The surface area of one cube with side length 18 cm is:

\( 6 \times (18 \, \text{cm} \times 18 \, \text{cm}) = 1944 \, \text{cm}^2 \).

- Therefore, the sum of the surface areas of the 8 cubes is:

\( 8 \times 1944 \, \text{cm}^2 = 15552 \, \text{cm}^2 \).

- Option 4 (15552) is the correct answer.