The conversion of A to B follows second order kinetics. Doubling the concentration of A will increase the rate of formation of B
by a factor

$\text{ Rate }=\mathrm{k}\times [\mathrm{A}{]}^2$
The conversion of reaction is$\mathrm{A}\to \mathrm{B}$
For the second-order reaction, Rate $=\mathrm{k}\times [\mathrm{A}{]}^2$
If the concentration of A is doubled, then A becomes $2 \mathrm{A}$ in the rate reaction, which implies,
$\text{ Rate }=\mathrm{k}\times [2 \mathrm{A}{]}^2$
$\text{ Rate }=4\times \mathrm{k}\times [\mathrm{A}{]}^2$
$\therefore$The rate of the formation of B will increase by a factor of 4 .

Therefore, the correct option is (A).