I did it in this step: First I use ΔTb=kbm And substitute following Tb=101.45 and kb=0.512 I get m=198.144 mol/kg

This should be the first clue that something went wrong. 198 mol of sucrose is about 67 kilograms of sucrose, so your implied molality means that for every kg of water, there are 67 kg of sucrose. That can't be right. (It would be about the same as adding a tablespoon or two of water to a 3-lb bag of sucrose and expecting the whole thing to dissovlve.)

I think you forgot to use the change in boiling point temperature in your $\Delta T_b = k_b m$ equation. The change is only 1.45 °C, not 101.45 °C.

What happens if you plug in that value and redo your calculations?

Update: Now that you've done that, your answer is at least chemically reasonable but there is still one problem.

Density equal to 1.036 g/ml equals 1.036 kg/L, which means 1.036 kg for 1 L and 1 kg=0.965 L.

This is true, but keep in mind the definition of molality: it is moles of solute per kg of solvent. However the density is obviously referring to the kg of total solution. So, it is inappropriate to find the molarity by simply dividing the molality by the density. As I described in a past answer on chem.se, if the molarity is $C$, the molality is $\frac{1000}{f}\frac{\frac{C f}{\rho}}{1-\frac{Cf}{\rho}}$, where $f$ is the formula weight of the solute and $\rho$ is the density. So you will have to invert this relationship to find the molarity from the molality.

I did it in this step: First I use ΔTb=kbm And substitute following Tb=101.45 and kb=0.512 I get m=198.144 mol/kg

This should be the first clue that something went wrong. 198 mol of sucrose is about 67 kilograms of sucrose, so your implied molality means that for every kg of water, there are 67 kg of sucrose. That can't be right. (It would be about the same as adding a tablespoon or two of water to a 3-lb bag of sucrose and expecting the whole thing to dissovlve.)

I think you forgot to use the change in boiling point temperature in your $\Delta T_b = k_b m$ equation. The change is only 1.45 °C, not 101.45 °C.

What happens if you plug in that value and redo your calculations?

Update: Now that you've done that, your answer is at least chemically reasonable but there is still one problem.

Density equal to 1.036 g/ml equals 1.036 kg/L, which means 1.036 kg for 1 L and 1 kg=0.965 L.

This is true, but keep in mind the definition of molality: it is moles of solute per kg of solvent. However the density is obviously referring to the kg of total solution. So, it is inappropriate to find the molarity by simply dividing the molality by the density. As I described in a past answer on chem.se, if the molarity is $C$, the molality is $\frac{1000}{f}\frac{\frac{C f}{\rho}}{1-\frac{Cf}{\rho}}$, where $f$ is the formula weight of the solute and $\rho$ is the density. So you will have to invert this relationship to find the molarity from the molality.

I did it in this step: First I use ΔTb=kbm And substitute following Tb=101.45 and kb=0.512 I get m=198.144 mol/kg

This should be the first clue that something went wrong. 198 mol of sucrose is about 67 kilograms of sucrose, so your implied molality means that for every kg of water, there are 67 kg of sucrose. That can't be right. (It would be about the same as adding a tablespoon or two of water to a 3-lb bag of sucrose and expecting the whole thing to dissovlve.)

I think you forgot to use the change in boiling point temperature in your $\Delta T_b = k_b m$ equation. The change is only 1.45 °C, not 101.45 °C.

What happens if you plug in that value and redo your calculations?

I did it in this step: First I use ΔTb=kbm And substitute following Tb=101.45 and kb=0.512 I get m=198.144 mol/kg

This should be the first clue that something went wrong. 198 mol of sucrose is about 67 kilograms of sucrose, so your implied molality means that for every kg of water, there are 67 kg of sucrose. That can't be right. (It would be about the same as adding a tablespoon or two of water to a 3-lb bag of sucrose and expecting the whole thing to dissovlve.)

I think you forgot to use the change in boiling point temperature in your $\Delta T_b = k_b m$ equation. The change is only 1.45 °C, not 101.45 °C.

What happens if you plug in that value and redo your calculations?

Update: Now that you've done that, your answer is at least chemically reasonable but there is still one problem.

Density equal to 1.036 g/ml equals 1.036 kg/L, which means 1.036 kg for 1 L and 1 kg=0.965 L.

This is true, but keep in mind the definition of molality: it is moles of solute per kg of solvent. However the density is obviously referring to the kg of total solution. So, it is inappropriate to find the molarity by simply dividing the molality by the density. As I described in a past answer on chem.se, if the molarity is $C$, the molality is $\frac{1000}{f}\frac{\frac{C f}{\rho}}{1-\frac{Cf}{\rho}}$, where $f$ is the formula weight of the solute and $\rho$ is the density. So you will have to invert this relationship to find the molarity from the molality.