This is the result:

The ratio is $3/7$.

Here’s my own reasoning:

We will use the unit square to generate the carpet. First, we calculate the number of squares that are removed from the unit square at the $k$-th iteration. Notice that once we have removed the first central square, we have other 8 squares left, so at the second iteration we will remove the central squares from the 8 subsquares. Each subsquare has been divided in 9 sub-subsquares and we have removed the central one, so at the third iteration we will remove $8\cdot8=64$ squares. In general we will remove $8^{k-1}$ squares at the $k$-th iteration. Let’s call the squares removed at the $k$-th iteration $k$-squares from now on.

Notice that each $k$-square has a side length of $1/3^k$, so their area is $9^{-k}$. Therefore the area of the carpet at the $k$-th iteration is $1-\sum_{n=0}^{k-1} 9^{-(n+1)}$

Without loss of generality, we can study only the lower-left quadrant of the carpet, where we want to find the number of squares removed from the carpet and outside of the red square, and then to get the total we will multiply this by four, since the quadrants are symmetric. Let’s call the sequence of these numbers $a_n$. Consider the second iteration: then $a_1=1$. If we count those at the third iteration we get $a_3=9$. In general, we may notice from the image that $a_n$ is the number of removed squares at the $k-1$-th iteration (due to the fractal’s autosimilarity to the children sub-carpets) minus the number of $k-1$-squares removed, which is $a_{k-2}$, plus two times $a_{k-2}$. These $k-1$-squares are marked with the blue color in the following image in the case $k=3$. So we get the following recurrence relation: $a_n=8^{n-1}+a_{n-1}$ where $n=k-1$.

The solution to this recurrence is $a_n=\frac{8^n-1}{7}$, and $4a_n$ is the total number of $k-1$-squares outside the red square. The area of the red square at the $k$-th iteration if we remove the areas of the squares in it will be $0.5-\frac{1}{9}\sum_{n=0}^{k-1}\left(8^n-4\cdot\frac{8^n-1}{7}\right)9^{-n}$

Now, the ratio we are looking for is just the ratio between the two areas at the $k$-th iteration, as $k$ goes to infinity. Let A be the area of the carpet, and B the area of the carpet restricted to the red square. The ratio will be $r=B/A=$ $$=\frac{0.5-\frac{1}{9}\sum_{n=0}^{k-1}\left(8^n-4\cdot\frac{8^n-1}{7}\right)9^{-n}}{1-\sum_{n=0}^{k-1} 9^{-(n+1)}}$$. The sums are just truncated geometric series, so after writing them explicitly, and after some algebra we get $r=3/7-\frac{1}{14\cdot8^{k+1}}$ and as $k\to\infty$ the ratio is just $$r=\frac{3}{7}$$