1. Given: All four bodies have the same mass $M$ and radius $2R$.

We need to evaluate: $2(I_2+I_3)+I_4=xI_1$

So first, write each moment of inertia.

2. Moment of inertia of solid sphere about its diameter $I_1=\frac{2}{5}M(2R)^2$ $I_1=\frac{2}{5}M\cdot 4R^2=\frac{8}{5}MR^2$

3. Moment of inertia of solid cylinder about its axis For a solid cylinder about its own axis, $I_2=\frac{1}{2}M(2R)^2$ $I_2=\frac{1}{2}M\cdot 4R^2=2MR^2$

4. Moment of inertia of solid disc about its diameter For a solid disc, M.I. about a diameter is $I_3=\frac{1}{4}M(2R)^2$ $I_3=\frac{1}{4}M\cdot 4R^2=MR^2$

5. Moment of inertia of thin circular ring about its diameter For a thin ring, M.I. about a diameter is $I_4=\frac{1}{2}M(2R)^2$ $I_4=\frac{1}{2}M\cdot 4R^2=2MR^2$

6. Substitute into the given expression $2(I_2+I_3)+I_4=2(2MR^2+MR^2)+2MR^2$ $=2(3MR^2)+2MR^2$ $=6MR^2+2MR^2=8MR^2$

So, $xI_1=8MR^2$

Using $I_1=\frac{8}{5}MR^2$, $x\left(\frac{8}{5}MR^2\right)=8MR^2$

Cancel $8MR^2$ from both sides: $\frac{x}{5}=1$ $x=5$

7. Final Answer $\boxed{5}$

The derived answer matches the stored correct answer.