Let $$A$$ be a matrix of order $$3 \times 3$$ and $$\det(A) = 2$$. Then $$\det(\det(A) \text{ adj}(5 \text{ adj}(A^3)))$$ is equal to

Since $$A$$ is a $$3 \times 3$$ matrix with $$\det(A)=2$$, we aim to determine $$\det(\det(A)\cdot\text{adj}(5\,\text{adj}(A^3)))$$. We first recall that for an $$n\times n$$ matrix $$M$$ (with $$n=3$$ here), the following hold: $$\det(kM)=k^n\det(M)$$, $$\det(\text{adj}(M))=(\det(M))^{n-1}$$, $$\text{adj}(kM)=k^{n-1}\,\text{adj}(M)$$, and $$\text{adj}(\text{adj}(M))=(\det(M))^{n-2}\cdot M$$.

Applying $$\text{adj}(kM)=k^{n-1}\,\text{adj}(M)$$ with $$k=5$$ and $$M=\text{adj}(A^3)$$ gives $$\text{adj}(5\,\text{adj}(A^3))=5^2\cdot\text{adj}(\text{adj}(A^3))=25\cdot\text{adj}(\text{adj}(A^3))$$. Then using $$\text{adj}(\text{adj}(M))=(\det(M))^{n-2}\cdot M$$ with $$M=A^3$$ yields $$\text{adj}(\text{adj}(A^3))=(\det(A^3))^{3-2}\cdot A^3=\det(A^3)\cdot A^3$$. Since $$\det(A^3)=(\det(A))^3=2^3=8$$, we obtain $$\text{adj}(\text{adj}(A^3))=8A^3$$ and hence $$\text{adj}(5\,\text{adj}(A^3))=25\times 8A^3=200A^3$$.

Therefore, $$\det(A)\cdot\text{adj}(5\,\text{adj}(A^3))=2\times 200A^3=400A^3$$, and its determinant is $$\det(400A^3)=400^3\cdot\det(A^3)=400^3\times 8$$. Noting that $$400^3=(4\times10^2)^3=64\times10^6$$, we find $$\det(400A^3)=64\times10^6\times8=512\times10^6$$.

Therefore the correct answer is Option C: $$512 \times 10^6$$.