Moment of inertia of the solid sphere about any diameter is given by, $\frac{2m_1{R}^2}{5}$.

Using parallel axis theorem, we get

$$\begin{gathered} I_{\text{solid sphere }}=\frac{2}{5}{m}_1{R}^2+m_1{R}^2 \\ =\frac{7}{5}{m}_1{R}^2 \end{gathered}$$

$=\frac{7}{5}\times 5R^2$

$=7R^2$

Moment of inertia of the disc about the diameter is given by, $\frac{m_2{R}^2}{4}$.

Now using parallel axis theorem, we get

$$\begin{gathered} I_{\text{disc }}=\frac{m_2{R}^2}{4}+m_2{R}^2 \\ =\frac{5}{4}{m}_2{R}^2 \end{gathered}$$

$=\frac{5}{4}\times 4R^2$

$=5R^2$

Therefore,$\frac{I_{\text{disc }}}{I_{\text{soild sphere }}}=\frac{5}{7}$

Thus, $x=5$.