Answer: Option B
:
B
Given that surface area of the original cube = 24 units.
$\therefore$ Length of each side of the cube =$\sqrt{\frac{24}{6}}$= 2 units.
The radius of the biggest possible sphere that can be kept inside the cube of each side 2 units.
= $\frac{2}{2}$ = 1 unit.
Now, the diagonal of the second cube inscribed in the sphere = Diameter of sphere.
$\Rightarrow$$\sqrt{3}$ (length of each side of second cube) = 2 units.
$\Rightarrow$Length of each side of second cube =$\frac{2}{\sqrt{3}}$ unit
$\therefore$ surface area of the inner cube = 6 x${\left(\frac{2}{\sqrt{3}}\right)}^2$
= $6\times \frac{4}{3}=$ 8 units