Cut Cube

Let the edge of larger cube be A and that of smaller cube be a, then by Volume conservation we get $A^3=8a^3$ or A=2a.
Surface area of larger cube $=6A^2=24a^2.$
Surface area of smaller 8 cubes $=8(6a^2)=48a^2.$

Now difference of surface areas= $48a^2 - 24a^2$ $=24a^2=2016$ $\Rightarrow a=\sqrt{84}$ Hence volume of larger cube =8( $\sqrt{84})^3$ = 6159...(approximately)

Consider the diagram. Let $S_1$ be the surface area of the original cube and $S_2$ be the surface area of the $8$ small cubes. Then $S_1=6(2a)^2=24a^2$ and $S_2=8(6a^2)=48a^2$ . From the problem, it states that

$S_2=S_1+2016$

Substituting, we have

$48a^2=24a^2+2016$

$a^2=84$

$a=\sqrt{84}=2\sqrt{21}$

The volume of the original cube is

$V=(2a)^3=(4\sqrt{21})^3 \approx$ $\boxed{6159}$