Answer
See the answer below.
Work Step by Step
$n_{NaCl}=200\ g\div 58.44\ g/mol=3.42\ mol; i_{NaCl}\approx2\rightarrow i.m=6.84$
$n_{CaCl_2}=200\ g\div 110.98\ g/mol=1.82\ mol; i_{CaCl_2}\approx3\rightarrow i.m=5.46$
NaCl will have a higher freezing point depression and is cheaper, so it should be chosen.
