Answer
$\det (B)=-18$
Work Step by Step
Let $A=\begin{bmatrix}
a & b\\
c&d
\end{bmatrix}$ and $\det (A)=ad-bc=1$
$\det (B)=\begin{bmatrix}
-6d & -6c\\
3b & 3a
\end{bmatrix}$
The determinant of B is:
$\det (B)=-18ad+18bc=-18(ad-bc)=-18\det (A)=-18.1=-18$
