1. Useful determinant facts for a $3\times 3$ matrix

For a matrix of order $3$:

• $\det(\operatorname{adj}(M)) = (\det M)^{3-1} = (\det M)^2$
• $\det(kM)=k^3\det(M)$
• $\det(M^{-1})=\frac{1}{\det(M)}$

Given: $\det(A)=3$

We need to evaluate $\det\left(\operatorname{adj}\left(-4 \, \operatorname{adj}\left(-3 \, \operatorname{adj}\left(3 \, \operatorname{adj}\left((2A)^{-1}\right)\right)\right)\right)\right)=2^m3^n$

2. Start from the innermost matrix

Let $M_1=(2A)^{-1}$ Then $\det(2A)=2^3\det(A)=8\cdot 3=24$ So $\det(M_1)=\det\big((2A)^{-1}\big)=\frac{1}{24}=2^{-3}3^{-1}$

3. First adjugate

Let $M_2=\operatorname{adj}(M_1)$ Then $\det(M_2)=\det(\operatorname{adj}(M_1))=(\det M_1)^2=\left(\frac{1}{24}\right)^2=2^{-6}3^{-2}$

Now multiply by $3$: $M_3=3M_2$ Hence $\det(M_3)=3^3\det(M_2)=27\cdot 2^{-6}3^{-2}=2^{-6}3^1$

4. Second adjugate

Let $M_4=\operatorname{adj}(M_3)$ Then $\det(M_4)=(\det M_3)^2=(2^{-6}3^1)^2=2^{-12}3^2$

Now multiply by $-3$: $M_5=-3M_4$ Since order is $3$, $\det(M_5)=(-3)^3\det(M_4)=-27\cdot 2^{-12}3^2=-2^{-12}3^5$

5. Third adjugate

Let $M_6=\operatorname{adj}(M_5)$ Then $\det(M_6)=(\det M_5)^2=(-2^{-12}3^5)^2=2^{-24}3^{10}$

Now multiply by $-4$: $M_7=-4M_6$ So $\det(M_7)=(-4)^3\det(M_6)=-64\cdot 2^{-24}3^{10}=-2^6\cdot 2^{-24}3^{10}=-2^{-18}3^{10}$

6. Final adjugate

Required matrix is $M_8=\operatorname{adj}(M_7)$ Therefore $\det(M_8)=(\det M_7)^2=(-2^{-18}3^{10})^2=2^{-36}3^{20}$

So, $2^m3^n=2^{-36}3^{20}$ which gives $m=-36,\quad n=20$

Hence $m+2n=-36+40=4$

7. Check with options

The value is $\boxed{4}$ So the correct option is B.