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$$\begin{gathered} q=2 \\ \stackrel{鈫�}{v}=2.0\stackrel{铃�}{i}+4.0\stackrel{铃�}{j}+6.0\stackrel{铃�}{k} \\ \stackrel{鈫�}{F}=4.0\stackrel{铃�}{i}-20\stackrel{铃�}{j}+12\stackrel{铃�}{k} \end{gathered}$$

Using the formula for cross-product we can find the vector $\stackrel{\mathbf{鈫�}}{\mathbf{B}}$from the given vectors$\stackrel{\mathbf{鈫�}}{\mathbf{F}}$and$\stackrel{\mathbf{鈫�}}{\mathbf{v}}$.

The vector product is written as,

$\left(\stackrel{鈫�}{a}脳\stackrel{鈬赌}{b}\right)=\left(a_y{b}_z-b_y{a}_z\right)\stackrel{铃�}{i}+\left(a_z{b}_x-b_z{a}_x\right)\stackrel{铃�}{j}+\left(a_x{b}_y-b_y{a}_x\right)\stackrel{铃�}{k}$ (i)

Use equation (i) to write the cross product between $\stackrel{鈫�}{v}$and $\stackrel{鈫�}{B}$.

$$\begin{gathered} \stackrel{鈫�}{F}=q\stackrel{鈫�}{v}脳\stackrel{鈫�}{B} \\ {F}_x\stackrel{铃�}{i}+F_y\stackrel{铃�}{j}+F_z\stackrel{铃�}{k}=q\left(v_y{B}_z-v_z{B}_y\right)\stackrel{铃�}{i}+q\left(v_z{B}_x-v_x{B}_z\right)\stackrel{铃�}{j}+q\left(v_x{B}_y-v_y{B}_x\right)\stackrel{铃�}{k} \end{gathered}$$

Comparing the above equation with the given vector$\stackrel{鈫�}{F}$, we get

$$\begin{gathered} 2\left(4B_z-6B_y\right)=4 \\ 2\left(6B_x-2B_z\right)=-2 \\ 2\left(2B_y-4B_x\right)=12 \end{gathered}$$

As $B_x=B_y$, above equation becomes,

$$\begin{gathered} 2\left(2B_y-4B_y\right)=12 \\ 4B_y-8B_y=12 \\ -4B_y=12 \\ {B}_y=-3 \end{gathered}$$

Inserting this value in above equation we get,

$$\begin{gathered} 2\left(4B_z-6\left(-3\right)\right)=4 \\ {B}_z=-4 \end{gathered}$$

Calculate the value of $B_x$from the above equations.

$$\begin{gathered} B_x=B_y \\ {B}_x=-3 \end{gathered}$$

Therefore, the vector$\stackrel{鈫�}{\mathrm{B}}$ can be written as$-3.0\stackrel{铃�}{i}-3.0\stackrel{铃�}{j}-4.0\stackrel{铃�}{k}$