Characteristic roots are nothing but eigenvalues. So the eigenvalues of matrix A are $3,2,-1$.
Also, $determinant\ of\ matrix\ B = Product\ of\ its\ eigenvalues.$
Now eigenvalues of matrix $B$ can be found by substituting the corresponding eigenvalues of matrix $A$.
$B= A^2-A$
$1.\ B = 3^2-3=9-3=6$
$2.\ B=2^2-2=4-2=2$
$3.\ B=-1^2-(-1)=1+1=2$
$|B|=6*2*2=24\ (Answer)$