Solution:

The answer is (B). Let $\Gamma$ be the initial circle and $C_{1}, C_{2}, \ldots, C_{44}$ the circles drawn centered at the vertices of the polygon labeled with perfect squares; note that, since the radius $R$ of each circle $C_{i}$ equals the radius of $\Gamma$ (all circles have perimeter $2019=2 \pi R$), $C_{i}$ passes through the center of $\Gamma$, which we will call $O$. We also note that for every $i$ the circles $C_{i}$ and $C_{i+1}$ (in the case $i=44$ the circles $C_{44}$ and $C_{1}$: from now on we index all elements cyclically, that is we set by convention $i+1=1$ for $i=44$) intersect at a point $P_{i}$ outside the circle $\Gamma$. This is because, denoting by $O_{i}$ the centers of the circles, the minimum value for the angle $\widehat{O_{i} O O_{i+1}}$ such that the intersection of $C_{i}$ and $C_{i+1}$ other than $O$ does not lie outside $\Gamma$ is such that $O O_{i}=O P_{i}=O O_{i+1}$ and it is equal to $2 \pi / 3$; however, the angle $\widehat{O_{i} O O_{i+1}}$ is in our case at most $\frac{(i+1)^{2}-i^{2}}{2019} \cdot 2 \pi=\frac{2 i+1}{2019} \cdot 2 \pi \leq \frac{89}{2019} \cdot 2 \pi<2 \pi / 3$.

The perimeter of the castle is thus obtained by summing, for $i$ ranging from $1$ to $44$, the length of the arc $P_{i} P_{i+1}$ of the circle $C_{i+1}$ that does not contain the point $O$; this length is equal to $R \theta_{i}$, where $\theta_{i}$ is the measure of the angle $P_{i}\widehat{O_{i+1} P}_{i+1}$ in radians. But, since they insist on the same arc of $C_{i+1}$, we have that the angle $P_{i} \widehat{O_{i+1} P_{i+1}}$ is double the angle $\widehat{P_{i} P_{i+1}}$; and, summing over all $i$ from $1$ to $44$, we obtain that $\theta_{1}+\ldots+\theta_{44}$ is thus equal to $4 \pi$ (the angles $\widehat{P_{i} O_{i+1}}$ cover precisely a full angle). The perimeter of the castle is thus $4 \pi R$ and, since we know that $2 \pi R$ equals $2019$ meters, the correct answer is $4038$ meters.