$N$ is an integer whose representation in base $b$ is $777$.  Find the smallest integer $b$ for which $N$ is the fourth power of an integer.
1. We start by expressing the number \( N \) in base \( b \). Given that \( N \) is represented as \( 777 \) in base \( b \), we can write:
   \[
   N = 7b^2 + 7b + 7
   \]
   We need to find the smallest integer \( b \) such that \( N \) is a fourth power of an integer, i.e., \( N = a^4 \) for some integer \( a \).

2. Let's assume \( a = 7k \) for some integer \( k \). Then:
   \[
   N = (7k)^4 = 7^4 k^4
   \]
   Therefore, we need:
   \[
   7b^2 + 7b + 7 = 7^4 k^4
   \]
   Dividing both sides by 7, we get:
   \[
   b^2 + b + 1 = 7^3 k^4
   \]

3. We need to find the smallest \( b \) such that \( b^2 + b + 1 \) is a multiple of \( 7^3 = 343 \). Let's denote:
   \[
   b^2 + b + 1 = 343m
   \]
   for some integer \( m \).

4. We need to find the smallest \( b \) such that \( b^2 + b + 1 \) is a multiple of 343. We can test values of \( b \) to find the smallest solution.

5. Let's start with \( b = 18 \):
   \[
   b^2 + b + 1 = 18^2 + 18 + 1 = 324 + 18 + 1 = 343
   \]
   Since \( 343 = 7^3 \), we have:
   \[
   b^2 + b + 1 = 343 \cdot 1
   \]
   This satisfies our condition.

6. Therefore, the smallest integer \( b \) for which \( N \) is the fourth power of an integer is \( b = 18 \).

The final answer is \(\boxed{18}\).