The correct answer is A. **Step-by-step Solution:** 1. **Understand the Principle:** When a solid object is completely submerged in a liquid, the volume of the liquid displaced is equal to the volume of the submerged object. In this case, the volume of the sphere is equal to the volume of the water that rises in the cylindrical vessel. 2. **Formulate Volumes:** * The volume of the solid metallic sphere ($V_{sphere}$) is given by the formula: $V_{sphere} = \frac{4}{3}\pi R^3$. * The volume of the water displaced in the cylindrical vessel ($V_{displaced}$) can be considered as a cylinder of radius $r$ and height $h$. So, $V_{displaced} = \pi r^2 h$. 3. **Equate the Volumes:** According to the principle from Step 1: $V_{sphere} = V_{displaced}$ $\frac{4}{3}\pi R^3 = \pi r^2 h$ 4. **Simplify the Equation:** We can cancel $\pi$ from both sides: $\frac{4}{3} R^3 = r^2 h$ 5. **Incorporate the Given Condition:** The problem states that the radius of the sphere is twice the height by which the water level rises, i.e., $R = 2h$. From this, we can express $h$ in terms of $R$: $h = \frac{R}{2}$. 6. **Substitute and Solve for the Ratio:** Substitute the expression for $h$ into the simplified equation from Step 4: $\frac{4}{3} R^3 = r^2 \left(\frac{R}{2}\right)$ To find the ratio $r:R$, we need to isolate $r^2$ and $R^2$. First, multiply both sides by 2: $2 \times \frac{4}{3} R^3 = r^2 R$ $\frac{8}{3} R^3 = r^2 R$ Now, divide both sides by $R$ (since $R$ is a radius, $R \neq 0$): $\frac{8}{3} R^2 = r^2$ We want the ratio $r:R$, which means we need $\frac{r}{R}$. Let's rearrange the equation: $\frac{r^2}{R^2} = \frac{8}{3}$ Take the square root of both sides (since radii are positive, we take the positive root): $\frac{r}{R} = \sqrt{\frac{8}{3}}$ 7. **Simplify the Radical:** $\frac{r}{R} = \frac{\sqrt{8}}{\sqrt{3}} = \frac{\sqrt{4 \times 2}}{\sqrt{3}} = \frac{2\sqrt{2}}{\sqrt{3}}$ To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$: $\frac{r}{R} = \frac{2\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{2\sqrt{6}}{3}$ Thus, the ratio of the radius of the cylinder to the radius of the sphere ($r:R$) is $\frac{2\sqrt{6}}{3}$. The final answer is $\boxed{\text{A}}$.