1. Moment of inertia of the disc about a tangent in its plane

For a disc, the moment of inertia about a diameter in its plane is $I_{\text{diameter}}=\frac{1}{4}MR^2$

Using the parallel axis theorem, the moment of inertia about a tangent in its plane is $I_{\text{disc,tangent}}=I_{\text{diameter}}+MR^2$ $=\frac{1}{4}MR^2+MR^2=\frac{5}{4}MR^2$

For the disc, $M=4\,\text{kg}$, so $I_{\text{disc,tangent}}=\frac{5}{4}(4)R^2=5R^2$

2. Moment of inertia of the solid sphere about a tangent

For a solid sphere, the moment of inertia about any diameter is $I_{\text{diameter}}=\frac{2}{5}MR^2$

Again using the parallel axis theorem for a tangent axis: $I_{\text{sphere,tangent}}=I_{\text{diameter}}+MR^2$ $=\frac{2}{5}MR^2+MR^2=\frac{7}{5}MR^2$

For the sphere, $M=5\,\text{kg}$, so $I_{\text{sphere,tangent}}=\frac{7}{5}(5)R^2=7R^2$

3. Required ratio

$\frac{I_{\text{disc,tangent}}}{I_{\text{sphere,tangent}}}=\frac{5R^2}{7R^2}=\frac{5}{7}$

Given that the ratio is $\frac{x}{7}$, we get $\frac{x}{7}=\frac{5}{7}$ So, $x=5$

4. Comparison with stored answer

Stored correct answer = $5$.

Our derived answer is also $5$, so it agrees.