Answer : The resulting concentrations of CV and NaOH are 0.0027 M and 0.025 M respectively.
Explanation :
Step 1 : Find moles of crystal violet and NaOH.
The molarity formula is

Molarity of crystal violet = 
The volume of crystal violet solution is 18 mL which is 0.018 L.
Moles of crystal violet = 
Moles of crystal violet = 5.4 x 10⁻⁵
Moles of NaOH = 
Moles of NaOH = 5.00 x 10⁻⁴
Step 2 : Find total volume of the solution
The total volume of the solution after mixing NaOH and crystal violet is
0.018 L + 0.00200 = 0.020 L
Step 3 : Use molarity formula to find final concentrations
Molarity of crystal violet = 
Final concentration of CV = 0.0027 M
Molarity of NaOH= 
NaOH is a strong base and dissociates completely as follows.

The mole ratio of NaOH and OH⁻ is 1:1 . Therefore the concentration of OH⁻ is same as that of NaOH.
Concentration of OH⁻ = 0.025 M