1. Use determinant and adjoint formulas for a $3\times 3$ matrix

For a matrix $M$ of order $3$:

$\det(\operatorname{adj} M)=\big(\det M\big)^{3-1}=(\det M)^2$

Also, for scalar $k$ and $3\times 3$ matrix $M$:

$\det(kM)=k^3\det(M)$

Let

$d=\det(A)\neq 0$

since $A$ is non-singular.

2. Simplify the inner matrix

We are given

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

Start with

$B=(\det A)A=dA$

Then

$\det(B)=\det(dA)=d^3\det(A)=d^4$

Now,

$\det(\operatorname{adj} B)=(\det B)^2=d^8$

So for

$C=2\operatorname{adj}(B)$

we get

$\det(C)=2^3\det(\operatorname{adj}B)=2^3d^8$

Now consider

$\operatorname{adj}(C)$

Then

$\det(\operatorname{adj} C)=(\det C)^2=(2^3d^8)^2=2^6d^{16}$

Finally,

$\det(3\operatorname{adj}C)=3^3\det(\operatorname{adj}C)=3^3\cdot 2^6 d^{16}$

This is given equal to

$3^{-13}2^{-10}$

Hence,

$3^3\cdot 2^6 d^{16}=3^{-13}2^{-10}$

So,

$d^{16}=3^{-16}2^{-16}=(6)^{-16}$

Therefore,

$d=\pm 6^{-1}=\pm \frac{1}{6}$

3. Now compute $\det(3\operatorname{adj}(2A))$

Let

$E=2A$

Then

$\det(E)=2^3\det(A)=8d$

So,

$\det(\operatorname{adj}E)=(\det E)^2=(8d)^2=64d^2$

Therefore,

$\det(3\operatorname{adj}(2A))=3^3\det(\operatorname{adj}(2A))=27\cdot 64 d^2$

Using $d^2=\left(\frac{1}{6}\right)^2=\frac{1}{36}$,

$\det(3\operatorname{adj}(2A))=27\cdot 64\cdot \frac{1}{36}$

$=\frac{27}{36}\cdot 64=\frac{3}{4}\cdot 64=48$

Now,

$48=2^4\cdot 3^1$

So,

$m=4,\qquad n=1$

4. Find the required value

$|3m+2n|=|3\cdot 4+2\cdot 1|=|12+2|=14$

5. Comparison with stored answer

Derived answer $=14$, which matches the stored correct answer.