\(S_{NaCl\left(25^oC\right)}=\dfrac{m_{NaCl}}{m_{H_2O}}.100=36\left(g\right)\\ \rightarrow m_{NaCl\left(tan\right)}=\dfrac{m_{H_2O}.S}{100}=\dfrac{150.36}{100}=54\left(g\right)\\ \rightarrow m_{dd}=m_{H_2O}+m_{NaCl}=54+150=204\left(g\right)\\ \rightarrow B\)