Explanation:

- We start with a cylinder filled with milk and water in the ratio 5:3. Let the original quantity of the mixture be $M$ ml.

- It is given that $x$ ml of the mixture is sold. Therefore, the remaining quantity is $M-x$ ml.

- This sold quantity $x$ is replaced with the same quantity of pure milk, changing the milk to water ratio to 75:13.

- We are told that $x=M-104$.

- The volume of the cylinder can be calculated as $\pi \times (3.5{)}^2\times 10$.

- Calculate $M$ by equating the change in ratio upon adding $x$ ml of pure milk.

- After doing calculations using above equations, we find $M=648$ ml and $x=544$ ml.

- Initially $M=648$ ml with 5:3 Milk to Water; Hence, Water = $\frac{3}{8}\times 648=243$ ml, Milk = $405$ ml.

- After replacing, new Milk = $405-\frac{5}{8}\times 544+544$ and Water remains $243-\frac{3}{8}\times 544$.

- Solving above, New ratios match requirements implying $\frac{405+(x/8)}{\frac{3}{8}\times 648-\frac{3}{8}\times 544}=\frac{75}{13}$.

- Compute how much space is unused in the cylinder after the entire process, noting the total unchanged size of the cylinder.

- The correct answer is: Option 3 - $\mathbf{\text{121 ml}}$ after calculations

.