I would say -16709 J, or -16.7 kJ.
Let's start with the balanced equation
C_12H_22O_11 + 12O_2 -> 12CO_2 + 11H_2O
The heat of reaction can be defined as the amount of heat the system gives off to its surroundings so that it can return to its initial temperature. The heat of reaction is the negative value of the heat absorbed by the calorimeter and its contents
q_(reaction) = -q_(calo r imeter), where
q_(cal o r i meter) = C * (chan g e i n temperature), C representing the heat capacity of the calorimeter (in this case, 4.9 (kJ)/(degree C );
We can determine that the change in temperature, defined as T_(f i nal) - T_(i n itial), is equal to 3.41 degrees.
Therefore, q_(cal o ri meter) = 4.9 (kJ)/(degree C) * 3.41 = 16.7 kJ, which determines the heat of combustion of sucrose to be
q_(sucrose) = - q_(cal o ri meter) = -16.7 kJ -> exothermic reaction
Now, since 1.01 grams of sucrose was burned, the heat of combustion of sucrose is approximately equal to -16.7 (kJ)/g. In other words, when 1 gram of sucrose is burned, 16.7 kJ of heat are released from the reaction.
For one mole we would have
q_(sucrose) = -16.7 (kJ)/(g) * 246 g/(mo l e) = -4108 (kJ)/(m o l e),
246 g/(m o l e) being sucrose's molar mass.
