Answer: C

Solution:

Only the inner $3 \times 3$ squares can start filled. For the center to be filled afterward, if it began empty it needs exactly $3$ filled neighbors, and if it began filled it needs $2$ or $3.$

Every other square must end empty. The key restriction is that no border square may acquire exactly three filled neighbors, which rules out filling all three squares along an outer edge of the $3 \times 3.$

If the center starts empty, exactly $3$ of its $8$ neighbors must be filled. Checking these triples up to square symmetry leaves four types. With ring coordinates centered at $(0,0),$ representatives are $$\begin{gathered} \{(-1,-1),(-1,0),(1,1)\},\\ \{(-1,-1),(-1,1),(1,-1)\},\\ \{(-1,-1),(-1,1),(1,0)\},\\ \{(-1,-1),(0,1),(1,0)\}. \end{gathered}$$ Their symmetry-orbit sizes are $8,4,4,4,$ giving $20$ configurations.

If the center starts filled, the same neighbor check leaves only the two configurations in which the other filled cells are opposite corner neighbors. Hence the total is $20+2=22.$

Thus, the correct answer is C.