1. We need to find $\left|A^T A\,(\operatorname{adj}(2A))^{-1}(\operatorname{adj}(4B))(\operatorname{adj}(AB))^{-1}AA^T\right|$ for $3\times 3$ matrices $A,B$ with $|A|=3,\qquad |B|=2.$

2. Use the determinant property for a product: $|PQRS\cdots|=|P|\,|Q|\,|R|\,|S|\cdots$ So, \begin{align*} &\left|A^T A,(\operatorname{adj}(2A))^{-1}(\operatorname{adj}(4B))(\operatorname{adj}(AB))^{-1}AA^T\right| \ &=|A^TA|\cdot |(\operatorname{adj}(2A))^{-1}|\cdot |\operatorname{adj}(4B)|\cdot |(\operatorname{adj}(AB))^{-1}|\cdot |AA^T|. \end{align*}

3. Compute the easy determinants first.

For any square matrix $M$, $|M^T|=|M|.$ Hence, $|A^TA|=|A^T||A|=|A|^2=3^2=9,$ and similarly, $|AA^T|=|A||A^T|=|A|^2=9.$

4. Use the adjoint determinant formula. For an $n\times n$ matrix $M$, $|\operatorname{adj}(M)|=|M|^{n-1}.$ Here $n=3$, so $|\operatorname{adj}(M)|=|M|^2.$ Therefore, $|(\operatorname{adj}(M))^{-1}|=\frac{1}{|\operatorname{adj}(M)|}=\frac{1}{|M|^2}.$

5. Compute each remaining factor.

(i) $|(\operatorname{adj}(2A))^{-1}|$

Since $A$ is $3\times 3$, $|2A|=2^3|A|=8\cdot 3=24.$ Thus, $|\operatorname{adj}(2A)|=|2A|^2=24^2,$ so $|(\operatorname{adj}(2A))^{-1}|=\frac{1}{24^2}.$

(ii) $|\operatorname{adj}(4B)|$

$|4B|=4^3|B|=64\cdot 2=128.$ Hence, $|\operatorname{adj}(4B)|=|4B|^2=128^2.$

(iii) $|(\operatorname{adj}(AB))^{-1}|$

First, $|AB|=|A||B|=3\cdot 2=6.$ So, $|\operatorname{adj}(AB)|=|AB|^2=6^2=36,$ and therefore, $|(\operatorname{adj}(AB))^{-1}|=\frac{1}{36}.$

6. Multiply all factors: \begin{align*} \text{Required determinant} &=9\cdot \frac{1}{24^2}\cdot 128^2\cdot \frac{1}{36}\cdot 9. \end{align*}

Now simplify: $9\cdot 9=81,$ so $=\frac{81\cdot 128^2}{24^2\cdot 36}.$

Since $128^2=16384,\qquad 24^2=576,$ we get $=\frac{81\cdot 16384}{576\cdot 36}.$

But a cleaner simplification is: $\frac{128^2}{24^2}=\left(\frac{128}{24}\right)^2=\left(\frac{16}{3}\right)^2=\frac{256}{9}.$ Thus, \begin{align*} \text{Required determinant} &=9\cdot \frac{256}{9}\cdot \frac{1}{36}\cdot 9 \ &=256\cdot \frac{9}{36} \ &=256\cdot \frac{1}{4} \ &=64. \end{align*}

7. Therefore the correct option is $\boxed{64}.$ So, Option C is correct.