On a windless day, a pigeon can fly from Albatrocity to Finchester and back in $3$ hours and $45$ minutes. However, when there is a $10$ mile per hour win blowing from Albatrocity to Finchester, it takes the pigeon $4$ hours to make the round trip. How many miles is it from Albatrocity to Finchester?
1. Let the distance from Albatrocity to Finchester be \( x \) miles.
2. Let the speed of the pigeon in still air be \( a \) miles per hour.

3. The total time for a round trip without wind is given as \( 3 \) hours and \( 45 \) minutes, which is \( 3.75 \) hours. Therefore, we have:
   \[
   \frac{2x}{a} = 3.75
   \]
   Solving for \( x \):
   \[
   2x = 3.75a \implies x = \frac{3.75a}{2} = \frac{15a}{8}
   \]

4. When there is a \( 10 \) mile per hour wind blowing from Albatrocity to Finchester, the pigeon’s effective speed going to Finchester is \( a + 10 \) miles per hour, and the effective speed returning to Albatrocity is \( a - 10 \) miles per hour. The total time for the round trip is \( 4 \) hours. Therefore, we have:
   \[
   \frac{x}{a+10} + \frac{x}{a-10} = 4
   \]

5. Substitute \( x = \frac{15a}{8} \) into the equation:
   \[
   \frac{\frac{15a}{8}}{a+10} + \frac{\frac{15a}{8}}{a-10} = 4
   \]
   Simplify the equation:
   \[
   \frac{15a}{8(a+10)} + \frac{15a}{8(a-10)} = 4
   \]
   Multiply through by \( 8(a+10)(a-10) \) to clear the denominators:
   \[
   15a(a-10) + 15a(a+10) = 32(a^2 - 100)
   \]
   Simplify:
   \[
   15a^2 - 150a + 15a^2 + 150a = 32a^2 - 3200
   \]
   Combine like terms:
   \[
   30a^2 = 32a^2 - 3200
   \]
   Rearrange the equation:
   \[
   32a^2 - 30a^2 = 3200
   \]
   Simplify:
   \[
   2a^2 = 3200
   \]
   Solve for \( a \):
   \[
   a^2 = 1600 \implies a = 40
   \]

6. Substitute \( a = 40 \) back into the expression for \( x \):
   \[
   x = \frac{15a}{8} = \frac{15 \times 40}{8} = 75
   \]

The final answer is \( \boxed{75} \) miles.