The freezing point of a liquid at a given pressure depends upon the ability of the molecules to stack on each other in a way where they bind to each other so they do not move. The main way in which an impurity can change this is by getting in the way of the molecules stacking on each other. For this reason, the larger the osmolality of the solution (the molality of particles of dissolved in the solution) the larger the freezing point is lowered. To put that into algebraic terms:
ΔT = Kf x m
Where m = molality = mol solute / kg solvent.
In the case of water, Kf = 1.86 (C° / m). However, the molality of PARTICLES is not the same as the molality of the SOLUTE, because some solutes ionize and become multiple moles of particles for every mole of solute. For example, NaCl ionizes in water to Na+ and Cl-, so 1.0mol of dissolved NaCl makes 2.0mol of dissolved ions. For the purpose of freezing point depression, we would go with the 2.0mol of ions when we calculate the molality of the solution.
In this question, all of the solutions have water as a solvent, which means that we can disregard the Kf value and instead concentrate upon the molality of solute particles of the solution. We can get that by first multiplying the moles of solute formula by the number of particles each formula unit of solute breaks up into when it dissolves, then dividing that by the mass of the solvent.
0.175mol KCl x (2mol ions / 1mol KCl) / 1.25kg solvent = 2.80m
0.175mol glucose x (1mol molecules / 1mol glucose) / 0.750kg solvent = 2.33m
0.075mol Ca(NO3)2 x (3mol ions / 1mol Ca(NO3)2) / 0.950kg solvent = 2.37m
The highest freezing point will be the one with the lowest molality of particles of solute, the glucose solution; the next highest will be the Ca(NO3)2 solution; and the lowest freezing point will be the KCl solution.