In a ten-mile race First beats Second by $2$ miles and First beats Third by $4$ miles. If the runners maintain constant speeds throughout the race, by how many miles does Second beat Third?

$ \textbf{(A)}\ 2\qquad\textbf{(B)}\ 2\frac{1}{4}\qquad\textbf{(C)}\ 2\frac{1}{2}\qquad\textbf{(D)}\ 2\frac{3}{4}\qquad\textbf{(E)}\ 3 $
1. Let's denote the speeds of First, Second, and Third as \( v_1 \), \( v_2 \), and \( v_3 \) respectively.
2. Given that First beats Second by 2 miles in a 10-mile race, we can write the relationship:
   \[
   \frac{10}{v_1} = \frac{8}{v_2}
   \]
   This implies:
   \[
   v_1 = \frac{10}{8} v_2 = \frac{5}{4} v_2
   \]
3. Similarly, given that First beats Third by 4 miles in a 10-mile race, we can write:
   \[
   \frac{10}{v_1} = \frac{6}{v_3}
   \]
   This implies:
   \[
   v_1 = \frac{10}{6} v_3 = \frac{5}{3} v_3
   \]
4. We need to find the distance by which Second beats Third. First, we express \( v_2 \) and \( v_3 \) in terms of \( v_1 \):
   \[
   v_2 = \frac{4}{5} v_1
   \]
   \[
   v_3 = \frac{3}{5} v_1
   \]
5. The time taken by Second to finish the race is:
   \[
   t_2 = \frac{10}{v_2} = \frac{10}{\frac{4}{5} v_1} = \frac{10 \cdot 5}{4 v_1} = \frac{50}{4 v_1} = \frac{25}{2 v_1}
   \]
6. The distance covered by Third in this time is:
   \[
   d_3 = v_3 \cdot t_2 = \left( \frac{3}{5} v_1 \right) \left( \frac{25}{2 v_1} \right) = \frac{3}{5} \cdot \frac{25}{2} = \frac{75}{10} = 7.5 \text{ miles}
   \]
7. Therefore, the distance by which Second beats Third is:
   \[
   10 - 7.5 = 2.5 \text{ miles}
   \]

The final answer is \(\boxed{2.5}\).