NaOH+ HCl--> NaCl+ H2O
From this balanced equation, we know that 1 mol NaOH= 1 mol HCl (keep in mind this because it will be used later).
We also know that 0.15 M HCl aqueous solution (soln)= 0.15 mol HCl/ 1 L of HCl soln (this one is based on the definition of molarity).
First, we should find the mole of HCl:
16.0 mL HCl soln* (1 L HCl soln/ 1,000 mL soln)* (0.15 mol HCl/ 1L HCl soln)= 2.40* 10^(-3) mol HCl.
Now, let's find the concentration of NaOH aqueous soln:
2.40* 10^(-3) mol HCl* (1 mol NaOH/ 1 mol HCl)* (1/ 12.0 mL NaOH soln)* (1,000 mL NaOH soln/ 1L NaOH soln)= 0.20 M NaOH aqueous soln.
The final answer is (2) 0.20 M NaOH (aq).
Also, this problem can also be done by using dimensional analysis. Don't forget significant figures.
Hope this would help~