A solid sphere of mass M and radius R is divided into two unequal...

Let the mass of the original solid sphere be $M$ and its radius be $R$.

Mass of the smaller sphere, $m_1 = \dfrac{M}{8}$.
Since the material remains the same, the density is constant. The volume is directly proportional to the mass.
$\dfrac{4}{3}\pi r^3 = \dfrac{1}{8} \left(\dfrac{4}{3}\pi R^3\right) \Rightarrow r^3 = \dfrac{R^3}{8} \Rightarrow r = \dfrac{R}{2}$

Moment of inertia of the smaller sphere about an axis through its centre is:
$I_1 = \dfrac{2}{5} m_1 r^2 = \dfrac{2}{5} \left(\dfrac{M}{8}\right) \left(\dfrac{R}{2}\right)^2 = \dfrac{2}{5} \times \dfrac{M}{8} \times \dfrac{R^2}{4} = \dfrac{M R^2}{80}$

Mass of the larger part (which is converted into a disc) is:
$m_2 = M - \dfrac{M}{8} = \dfrac{7M}{8}$

The radius of the disc is given as $2R$. The moment of inertia of a disc about its diameter is:
$I_2 = \dfrac{1}{4} m_2 (\text{radius})^2 = \dfrac{1}{4} \left(\dfrac{7M}{8}\right) (2R)^2 = \dfrac{1}{4} \times \dfrac{7M}{8} \times 4R^2 = \dfrac{7M R^2}{8}$

The ratio of their moments of inertia is:
$\dfrac{I_2}{I_1} = \dfrac{\dfrac{7M R^2}{8}}{\dfrac{M R^2}{80}} = \dfrac{7}{8} \times 80 = 70$

Answer: $70$