[THOUGHT] To solve this problem, we need to determine the maximum size of the balls of radius 1 in $\mathbb{Z}_2^{100}$ centered at any vector $\mathbf{x} \in \mathbb{Z}_2^{100}$.

1. Understanding the Fixed Length Levenshtein Distance: The Fixed Length Levenshtein distance between two vectors $\mathbf{u}$ and $\mathbf{v}$ in $\mathbb{Z}_q^n$ is defined as the smallest number $t$ such that $\mathbf{u}$ can be transformed into $\mathbf{v}$ through $t$ deletions and $t$ insertions. In the context of $\mathbb{Z}_2^{100}$, deletions and insertions are operations that modify the vector while keeping its length fixed.

2. Radius 1 Ball: A ball of radius 1 centered at $\mathbf{x}$ includes all vectors $\mathbf{y}$ such that the Fixed Length Levenshtein distance between $\mathbf{x}$ and $\mathbf{y}$ is at most 1. This means $\mathbf{y}$ can be obtained from $\mathbf{x}$ by performing at most 1 deletion and 1 insertion.

3. Calculating the Size of the Ball: For a vector $\mathbf{x} \in \mathbb{Z}_2^{100}$, the number of vectors $\mathbf{y}$ at a Fixed Length Levenshtein distance of 1 can be determined by considering all possible single deletions and insertions. However, since the vectors are in $\mathbb{Z}_2^{100}$, each component is either 0 or 1, and the operations are constrained.

4. Maximizing the Ball Size: The maximum size of the ball occurs when the center vector $\mathbf{x}$ allows the most distinct vectors $\mathbf{y}$ to be obtained through 1 deletion and 1 insertion. This is maximized when $\mathbf{x}$ has a balanced number of 0s and 1s, allowing the most flexibility in modifications.

5. Final Calculation: The size of the ball of radius 1 is given by the sum of the number of possible vectors obtained by 0 or 1 deletion and insertion operations. For $\mathbb{Z}_2^{100}$, this can be computed combinatorially.

[/THOUGHT] [PYTHON] from math import comb

n = 100

The maximum size occurs when the vector has 50 ones and 50 zeros.

k = n // 2

For each deletion of a 0 and insertion of a 0 (or similar for 1), the number of distinct vectors is maximized.

The total number of vectors is the sum of:

1. The original vector itself (distance 0).

2. Vectors obtained by deleting one 0 and inserting a 1 (or vice versa), which is 2 * k * (n - k).

3. Vectors obtained by deleting and inserting the same bit, which is n.

max_size = 1 + 2 * k * (n - k) + n print(max_size) [/PYTHON]

[THOUGHT] The Python output confirms that the maximum size of the ball of radius 1 in $\mathbb{Z}_2^{100}$ is 5101. [/THOUGHT] $\boxed{5101}$