The boiling point elevation of an aqueous solution can be determined using the formula:

$ \Delta \mathrm{T}_{\mathrm{b}} = i \mathrm{~K}_{\mathrm{b}} \times \mathrm{m} $

Where:

$ \Delta \mathrm{T}_{\mathrm{b}} $ is the change in boiling point.

$ i $ is the van’t Hoff factor (number of particles the solute breaks into).

$ \mathrm{K}_{\mathrm{b}} $ is the ebullioscopic constant of the solvent.

$ \mathrm{m} $ is the molality of the solution.

For simplicity, we'll assume molarity equals molality. Here's the calculation for each solution:

0.01 M Na$_2$SO$_4$:

$ i = 3 $ (Na$_2$SO$_4$ dissociates into 2 Na$^+$ and 1 SO$_4^{2-}$)

$ i \times m = 3 \times 0.01 = 0.03 $

0.015 M C$_6$HundefinedO$_6$ (glucose):

$ i = 1 $ (glucose does not dissociate)

$ i \times m = 1 \times 0.015 = 0.015 $

0.01 M Urea:

$ i = 1 $ (urea does not dissociate)

$ i \times m = 1 \times 0.01 = 0.01 $

0.01 M KNO$_3$:

$ i = 2 $ (KNO$_3$ dissociates into 1 K$^+$ and 1 NO$_3^-$)

$ i \times m = 2 \times 0.01 = 0.02 $

The boiling point elevation is directly proportional to $ i \times m $. Thus, the solution with the highest $ i \times m $ value will have the highest boiling point.

Therefore, the 0.01 M Na$_2$SO$_4$ solution, with an $ i \times m $ value of 0.03, will have the highest boiling point.