Farmer Boso has a busy farm with lots of animals. He tends to $5b$ cows, $5a +7$ chickens, and $b^{a-5}$ insects. Note that each insect has $6$ legs. The number of cows is equal to the number of insects. The total number of legs present amongst his animals can be expressed as $\overline{LLL }+1$, where $L$ stands for a digit. Find $L$.
1. We are given that the number of cows is equal to the number of insects. This can be expressed as:
   \[
   5b = b^{a-5}
   \]
   Dividing both sides by \( b \) (assuming \( b \neq 0 \)):
   \[
   5 = b^{a-6}
   \]
   Since \( a \) and \( b \) are integers, we need to find integer values of \( a \) and \( b \) that satisfy this equation. 

2. We test possible values of \( b \):
   - If \( b = 5 \):
     \[
     5 = 5^{a-6}
     \]
     This implies:
     \[
     5 = 5^{a-6} \implies a-6 = 1 \implies a = 7
     \]
     Therefore, \( b = 5 \) and \( a = 7 \) are solutions.

3. Now, we calculate the number of each type of animal:
   - Number of cows:
     \[
     5b = 5 \cdot 5 = 25
     \]
   - Number of chickens:
     \[
     5a + 7 = 5 \cdot 7 + 7 = 35 + 7 = 42
     \]
   - Number of insects:
     \[
     b^{a-5} = 5^{7-5} = 5^2 = 25
     \]

4. Next, we calculate the total number of legs:
   - Cows have 4 legs each:
     \[
     25 \text{ cows} \times 4 \text{ legs/cow} = 100 \text{ legs}
     \]
   - Chickens have 2 legs each:
     \[
     42 \text{ chickens} \times 2 \text{ legs/chicken} = 84 \text{ legs}
     \]
   - Insects have 6 legs each:
     \[
     25 \text{ insects} \times 6 \text{ legs/insect} = 150 \text{ legs}
     \]

5. Summing up all the legs:
   \[
   100 + 84 + 150 = 334
   \]
   We are given that the total number of legs can be expressed as \( \overline{LLL} + 1 \). Therefore:
   \[
   \overline{LLL} + 1 = 334 \implies \overline{LLL} = 333
   \]

6. From this, we can deduce that \( L = 3 \).

The final answer is \( \boxed{3} \)