Given:
- Weight of the block of wood = 2 kN
- Length of the inclined plane = 10 m
- Angle of the inclined plane with the horizontal = 15 degrees

To find:
- Work done in pulling up the block of wood

Assumptions:
- The inclined plane is smooth, meaning there is no friction between the block and the plane.
- The force applied is parallel to the incline, meaning it acts along the direction of the incline.

Solution:
Step 1: Resolving the weight of the block:
- The weight of the block can be resolved into two components: one perpendicular to the inclined plane and one parallel to the inclined plane.
- The component perpendicular to the inclined plane is mg * cos(theta), where m is the mass of the block (given as 2 kN) and theta is the angle of the inclined plane (given as 15 degrees).
- The component parallel to the inclined plane is mg * sin(theta).

Step 2: Calculating the work done:
- The work done in moving an object is given by the formula: Work = Force * Distance.
- In this case, the force applied is the component of weight parallel to the inclined plane, which is mg * sin(theta).
- The distance is the length of the inclined plane, which is given as 10 m.

Step 3: Calculation:
- The weight of the block can be calculated as mg = 2 kN.
- The angle theta is given as 15 degrees.
- The component of weight parallel to the inclined plane is mg * sin(theta) = 2 * sin(15) kN.
- The work done is given by the formula: Work = Force * Distance = (2 * sin(15)) kN * 10 m.
- Calculating this gives the answer as 5.17 kJ.

Therefore, the work done in slowly pulling up the block of wood for a length of 10 m on a smooth inclined plane at an angle of 15 degrees with the horizontal by a force parallel to the incline is 5.17 kJ.