Usage of all small properties

Let $2A=B$

$\Rightarrow |B|=|2A|=2^{3} |A|=8\times 5=40$ [Since $A$ is a $3 \times 3$ matrix]

Now, $B^{-1}=\frac{Adj(B)}{|B|}$

$\Rightarrow \big||B| B^{-1}\big|=|Adj(B)|$

$\Rightarrow |B|^{3}\frac{1}{|B|}=|Adj(B)|$

$\Rightarrow |B|^{2}=|Adj(B)|$

$\Rightarrow 40^{2}=|Adj(B)|$

$\Rightarrow 1600=|Adj(B)|$

$\Rightarrow 1600\times 27=3^{3} |Adj(B)|$

$\Rightarrow |3 Adj(B)|=\boxed{43200}$