A rectanglular block of dimensions 6m × 4m × 2m and of density 1.5 gm/c.c is lying on horizontal ground with the face of largest area in contact with the ground. The work done in arranging it with its smallest area in contact with the ground is, (g=10ms^–2)

We know that 

$mass=volume\times density$

substituting the values,

$$\begin{gathered} m=6\times 4\times 3\times 1.5\times {10}^3 \\ m=72\times {10}^{3}kg \end{gathered}$$

When a face with a big area touches the ground, its height is 2 metres.
While the face with the tiny area is in contact with the ground, the centre of mass is at a height of 1 m, and its height is 6 m.
The centre of mass is 3 metres high.

$$\begin{gathered} W=72\times {10}^4(3-1) \\ W=1440kJ \end{gathered}$$

Hence the correct answer is 1440kJ.