the rate of the reaction?

CH₃Br + NaOH → CH₃OH + NaBr
As the reaction is single-step bimolecular, the rate determination step is as same as the chemical equation, and
the rate law is: Rate = k [CH₃Br] [NaOH]

When the concentrations of CH3Br and NaOH are both 0.165 M , the rate of the reaction is 0.0010 M/s
0.0010 = k (0.165) (0.165) …… {1}

When concentration of CH₃Br is doubled:
Rate = k (0.165×2) (0.165) …… {2}
(2)/{1}: Rate/0.0010 = 2
Rate = 0.0020 M/s

When the concentrations of CH₃Br and NaOH are both increased by a factor of 3:
Rate = k (0.165×3) (0.165×3) …… {3}
{3}/{1}: Rate/0.0010 = 9
Rate = 0.0090 M/s