In the product
$\overrightarrow{\mathrm{F}}=\mathrm{q}(\overrightarrow{\mathrm{v}}\times \overrightarrow{\mathrm{B}})=\mathrm{q}\overrightarrow{\mathrm{v}}\times \left(\mathrm{B}\widehat{\mathrm{i}}+\mathrm{B}\widehat{\mathrm{j}}+{\mathrm{B}}_0\hat{\mathrm{k}}\right)$
For q = 1 and $\overrightarrow{\mathrm{v}}=2\widehat{\mathrm{i}}+4\widehat{\mathrm{j}}+6\widehat{\mathrm{k}}\text{ and }\overrightarrow{\mathrm{F}}=4\widehat{\mathrm{i}}-20\widehat{\mathrm{j}}+12\hat{\mathrm{k}}$
What will be the complete expression for $\overrightarrow{\mathrm{B}}$ ?

The correct option is:
$-6\widehat{i}-6\widehat{j}-8\hat{k}$

Step-by-Step Derivation and Explanation:

We are given the relation for the magnetic force acting on a moving charge:
$\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})$

Here, the magnetic field is expressed as:
$\overrightarrow{B}=B\widehat{i}+B\widehat{j}+B_0\hat{k}$

We are given the following values:
Charge: $q=1$
Velocity vector: $\overrightarrow{v}=2\widehat{i}+4\widehat{j}+6\widehat{k}$
Force vector: $\overrightarrow{F}=4\widehat{i}-20\widehat{j}+12\widehat{k}$

Substituting $q=1$ , the force equation becomes:
$\overrightarrow{F}=\overrightarrow{v}\times \overrightarrow{B}$

We calculate the cross product $\overrightarrow{v}\times \overrightarrow{B}$ using the determinant method:
$\overrightarrow{v}\times \overrightarrow{B}=\left|\begin{array}{ccc}\widehat{i}& \widehat{j}& \widehat{k}\\ 2& 4& 6\\ B& B& B_0\end{array}\right|$

Expanding this determinant, we get:
$\overrightarrow{v}\times \overrightarrow{B}=\widehat{i}(4B_0-6B)-\widehat{j}(2B_0-6B)+\widehat{k}(2B-4B)$
Simplifying the terms:
$\overrightarrow{v}\times \overrightarrow{B}=(4B_0-6B)\widehat{i}+(6B-2B_0)\widehat{j}-2B\widehat{k}$

We match the components of this vector with the components of the given force vector $\overrightarrow{F}=4\widehat{i}-20\widehat{j}+12\widehat{k}$:

1) From the $\widehat{k}$ component:
$-2B=12\Rightarrow B=-6$

2) From the $\widehat{i}$ component:
$4B_0-6B=4$
Substitute $B=-6$:
$4B_0-6(-6)=4$
$4B_0+36=4$
$4B_0=4-36=-32$
$B_0=-8$

3) Let us check the consistency of our calculated values using the $\widehat{j}$ component:
$6B-2B_0=6(-6)-2(-8)=-36+16=-20$
This perfectly matches the $\widehat{j}$ component of the given force, verifying our values.

Thus, the magnetic field vector is:
$\overrightarrow{B}=-6\widehat{i}-6\widehat{j}-8\hat{k}$