[THOUGHT] To determine the maximum size of the balls of radius 1 in $\mathbb{Z}_2^{100}$ under the Fixed Length Levenshtein distance, we consider the operations allowed: one deletion and one insertion. Each operation is defined by a deletion position $i$, an insertion position $j$, and an inserted bit $b \in \{0, 1\}$.

For a vector $\mathbf{x} \in \mathbb{Z}_2^{100}$, the ball of radius 1 includes:

1. The original vector $\mathbf{x}$ (t=0).
2. All vectors reachable by one deletion and one insertion (t=1).

The total number of possible operations for t=1 is $100 \times 100 \times 2 = 20,000$. However, some operations may produce the same vector. The maximum size occurs when overlaps are minimized.

Key observations:

• Deleting position $i$ and inserting $b$ at position $j=i$ is equivalent to flipping the bit at position $i$. There are $100 \times 2 = 200$ such operations, but they produce $100$ distinct vectors (each bit flip).
• For $j \neq i$, inserting $b$ at position $j$ after deleting $i$ can produce unique vectors if $\mathbf{x}$ is chosen such that no two operations yield the same result. This contributes $100 \times 99 \times 2 = 19,800$ operations. However, some of these may overlap with the bit flips or each other.

To maximize the number of distinct vectors, we assume that each operation (except those producing the original vector) results in a unique vector. The original vector is produced $100$ times (deleting $i$ and inserting the original bit at $i$). Thus, the number of distinct vectors from t=1 operations is $20,000 - 100 = 19,900$. Adding the original vector gives a total size of $19,900 + 1 = 19,901$.

However, upon closer analysis, the maximum distinct vectors from t=1 operations are actually $100 \times 100 \times 2 - 100 = 19,900$ (subtracting overlaps where the original vector is reproduced). Adding the original vector results in a ball size of $19,901$.

But considering the structure of $\mathbb{Z}_2^{100}$ and the nature of deletions/insertions, the correct maximum size is derived by recognizing that each deletion and insertion can uniquely alter the vector when the center $\mathbf{x}$ is chosen to minimize overlaps. The formula for the maximum size is $100 \times 100 \times 2 - 100 + 1 = 19,901$.

[PYTHON]

The calculation is based on the reasoning above.

The maximum size is 1001002 (all possible operations) minus 100 (overlaps producing the original vector) plus 1 (the original vector itself).

max_size = 100 * 100 * 2 - 100 + 1 print(max_size) [/PYTHON]

$\boxed{19901}$

[THOUGHT] The Python output confirms the calculation: 1001002 (all possible operations) minus 100 (overlaps producing the original vector) plus 1 (the original vector itself) yields 19,901. This is the maximum distinct vectors achievable under the Fixed Length Levenshtein distance with radius 1. $\boxed{19901}$