$$|\operatorname{adj}(2 \operatorname{adj}(|A| A))|=3^{-13} \cdot 2^{-10}$$

Let $$|A| A=B \Rightarrow|B|=\| A|A|=|A|^3|A|=|A|^4$$

$$\begin{aligned} \Rightarrow \quad & \operatorname{adj}(|A| A)=(\operatorname{adj} B) \\ \Rightarrow \quad & 2 \operatorname{adj}(|A| A)=(2 \operatorname{adj} B)=C \text { (say) } \\ & |\operatorname{3adj}(C)|=3^3 \cdot|C|^2 \end{aligned}$$

$$\begin{aligned} & |C|=|(2 \operatorname{adj} B)|=2^3|B|^2=2^3 \cdot\left|A^4\right|^2=2^3 \cdot|A|^8 \\ & \Rightarrow|\operatorname{3adj} C|=3^3 \cdot\left(2^3|A|^8\right)^2=3^{-13} \cdot 2^{-10} \\ & \quad=2^6|A|^{16}=3^{-16} \cdot 2^{-10} \\ & \Rightarrow|A|^{16}=(3 \cdot 2)^{-16}=\left(\frac{1}{6}\right)^{16} \\ & \Rightarrow|A|= \pm \frac{1}{6} \end{aligned}$$

$$\begin{array}{r} \mid \text { 3adj }\left.2 A\left|=3^3\right| 2 A\right|^2=3^3 \cdot\left(2^3|A|\right)^2=3^3 \cdot 2^6|A|^2 \\ =3^3 \cdot 2^6 \cdot \frac{1}{36}=\frac{27 \times 64}{36}=48 \end{array}$$

$$ \begin{aligned} & \Rightarrow 2^m \cdot 3^n=2^4 \cdot 3^1 \Rightarrow m=4 \\ & \qquad n=1 \\ & \Rightarrow|3 \times 4+2 \times 1|=14 \\ \end{aligned}$$