${\mathrm{CH}}_3-\mathrm{Br}\frac{\mathrm{Mg}}{\text{ ether }}\to \mathrm{X}\stackrel{{\left({\mathrm{CH}}_3\right)}_3\mathrm{C}-\mathrm{OH}}{\to }\mathrm{A}$. Product A is

The reaction can be represented as follows.

${\mathrm{CH}}_3-\mathrm{Br}\underset{\mathrm{ether}}{\overset{\mathrm{Mg}}{\to }}\underset{\mathrm{X}}{\mathrm{C}}{\mathrm{H}}_3-\mathrm{MgBr}\stackrel{{\left({\mathrm{CH}}_3\right)}_3\mathrm{C}-\mathrm{OH}}{\to }\underset{\mathrm{A}}{\mathrm{C}}{\mathrm{H}}_4$

In the above reaction the final product A formed is methane $\left(C{H}_4\right)$.

Hence, the correct answer is option (A) ${\mathrm{CH}}_4$.