[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}