Three cubes each having volume 1728 $$cm^{3}$$ are placed one above the other. What is the total surface area of the resulting solid?

Given,

Volume of Each cube = $$1728\ cm^3$$

Volume of cube = $$\left(side\right)^3$$

$$\left(side\right)^3=1728$$

$$\left(side\right)=\ 12$$

When three cubes are placed one over another it formed a Cuboid as shown in figure. 

Now, l =12 , b =12 , h = 36

Volume of Cuboid = 2(lb + bh + hl)

= 2($$\left(12\times\ 12\right)+\left(12\times\ 36\right)+\left(36\times\ 12\right)$$)

= 2 (144 + 432 + 432)

= 2 ( 1008 )

= 2016 $$cm^3$$

Hence, Option D is correct.