Emptied cube

Volume of the sculpture

Consider the sculpture to consist of three layers, each of height $1$ unit. Then the volumes of the bottom, middle and top layers are $5$ cube units, $2$ cube units, and $5$ cube units respectively. So the volume of the sculpture is $12$ cube units.

Alternatively: the sculpture consists of a $3\times 3\times 3$ cube from which two $2\times 2\times 2$ cubes have been removed. The $2\times 2\times 2$ cubes have exactly one $1\times 1\times 1$ cube in common - the cube at the centre of the $3\times 3\times 3$ cube. So the volume of the sculpture is

$$27\;\mathrm{units}^3-\left(2\times 8\;\mathrm{units}^3- 1\;\mathrm{units}^3\right) =12\;\mathrm{units}^3\,.$$

Image

Surface area of the sculpture

In each of six possible directions, the view of the sculpture is as sown,

with the outer square having side $3$ units and the inner square having

side $1$ unit. So the surface area of the sculpture is

$$6\times 8\;\mathrm{units}^2 = 48\;\mathrm{units}^2\;.$$