Let $vec(A)$ denote the vectorization operator, and let $\otimes$ denote the Kronecker product. We can then rewrite both sides of the equation to get $$ (I - B \otimes B) \,vec(A) = vec(CC^T) $$ Assuming invertibility, we have $$ vec(A) = (I - B \otimes B)^{-1}vec(CC^T) $$