$4\hat{ı}-20\hat{ȷ}+12k=(2\hat{ı}+4\hat{ȷ}+6\hat{k})\times \left(B\hat{ı}+B\hat{ȷ}+B_o\hat{k}\right)$

$4\hat{ı}-20\hat{ȷ}+12\hat{k}=\left|\begin{array}{ccc}+\hat{ı}& -\hat{ȷ}& +\hat{k}\\ 2& 4& 6\\ B& B& B_0\end{array}\right|$

$=\left(4B_0-6B\right)\hat{ı}-\left(2B_0-6B\right)\hat{ȷ}+(2B-4B)\hat{k}$

$4B_0-6B=4$

$2B_0-6B=20$

$2B-4B=12\Rightarrow B=6$

So, $2B_0-6(-6)=20$

$2B_0=20-36\Rightarrow B_0=-8$

$\therefore \overrightarrow{B}=-6\hat{ı}-6\hat{ȷ}-8\hat{k}$