$\mathrm{A}+2\mathrm{B}⟶\mathrm{C}$ the rate equation for this reaction is given as

Rate $=k\left[\mathrm{A}\right]\left[\mathrm{B}\right]$

If the concentration of A is kept the same but that of B is doubled what will happen to the rate itelf?

The rate law formulation will become Rate$=\mathrm{k} \left[\mathrm{B}\right]$ when the concentration of A is kept constant.
The rate of the reaction doubles with a doubling of B concentration.
New Rate$=k\times 2 \left[\mathrm{B}\right]=2\times$old rate.