If the length of the longest rod that can be placed in a cubical box is $18\sqrt{3} cm$, then what is the total surface area of the cube?

Let the side of the cube be a.

It is evident that the longest rod that can be placed in the cube is OC.

It is observed that: ΔOAB and ΔOBC are right-angled triangles.

Applying Pythagoras theorem in ΔOAB:

OB² = OA² + AB²

= a² + a²

= 2a²

Again applying Pythagoras theorem in ΔOBC:

OC² = OB² + BC²

= 2a² + a²

= 3a²

∴OC = $\sqrt{3}a$

Therefore, the longest rod that can be placed in a cube of side a is$\sqrt{3}a$.

$\sqrt{3}a=18\sqrt{3}$

⇒ a = 18 cm

Thus, total surface area of the cube = 6a² = 6 × (18 cm)² = 1944 cm².