If  A is a square matrix of order less than 4 such that  $|A-A^T|\ne 0$, and  $B=adjA$, then  $adj\left(B^2{A}^{-1}{B}^{-1}A\right)$ is

$A-A^T$  is skew symmetric matrix
Given that $|A-A^T|\ne 0$ 
$\therefore$  order is even and less than 4
$\therefore$ x = 2
B =  Adj A
$AB=A.AdjA\Rightarrow BA=Adj.A=\left|A\right|I$

$$\begin{gathered} \therefore AB=BA=\left|A\right|I \\ AB = BA \\ {B}^{-1}{A}^{-1}=A^{-1}{B}^{-1} \\ Adj\left(BB{A}^{-1}{B}^{-1}A\right) \\ Adj\left(BB{B}^{-1}{A}^{-1}A\right) \\ Adj\left(B\right) = A \end{gathered}$$