c
We know that,
$(a d j A)=A=|A| I=\left[\begin{array}{ccc}|A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A|\end{array}\right]$
$\Rightarrow|(a d j A) A|=\left[\begin{array}{ccc}|A| & 0 & 0 \\ 0 & |A| & 0 \\ 0 & 0 & |A|\end{array}\right]$
$\Rightarrow|\operatorname{adj} A||A|=|A|^{3}\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|=|A|^{3}(I)$
$\therefore|a d j A|=|A|^{2}$
Hence, the correct answer is $\mathrm{C}$.