since $A,\, B,\,C$ are all square matrices of order $2,$ and $C D-A B$ is well defined, $D$ must be a square matrix of order $2$ .

Let    $D=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] .$ Then $\mathrm{CD}-\mathrm{AB}=0$ gives

$\left[\begin{array}{ll}2 & 5 \\ 3 & 8\end{array}\right]\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]-\left[\begin{array}{cc}2 & -1 \\ 3 & 4\end{array}\right]\left[\begin{array}{ll}5 & 2 \\ 7 & 4\end{array}\right]=0$

or          $\left[\begin{array}{cc}2 a+5 c & 2 b+5 d \\ 3 a+8 c & 3 b+8 d\end{array}\right]-\left[\begin{array}{cc}3 & 0 \\ 43 & 22\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

or          $\left[\begin{array}{cc}2 a+5 c-3 & 2 b+5 d \\ 3 a+8 c-43 & 3 b+8 d-22\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

By equality of matrices, we get

$2 a+5 c-3=0$                  ……….. $(1)$

$3 a+8 c-43 =0$                 ……….. $(2)$

$2 b+5 d =0$                 ……….. $(3)$

and    $3 b+8 d-22 =0$                 ……….. $(4)$

Solving $(1)$ and $(2),$ we get $a=-191,\, c=77 .$ Solving $( 3)$ and $(4)$ we get $b=-110,$ $d=44$.

Therefore  $D=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]=\left[\begin{array}{cc}-191 & -110 \\ 77 & 44\end{array}\right]$