**Problem Statement:**
A block of mass 1 kg is placed on a wedge as shown in the figure. The surfaces are frictionless. The wedge is given a horizontal acceleration of 10 m/s^2 by applying a force on it such that the block does not slip on the wedge. We need to find the work done by the normal force in the ground frame in under root 3 seconds.

**Solution:**
Let's consider the motion of the block and wedge separately.

**Motion of the wedge:**
The force applied on the wedge is the only external force acting on it. Therefore, the acceleration of the wedge can be calculated using Newton's second law as follows:

F = ma
where F is the applied force, m is the mass of the wedge, and a is the acceleration of the wedge.

Here, the mass of the wedge is not given in the problem statement. However, it is not required to solve the problem as we only need to find the work done by the normal force on the block.

**Motion of the block:**
The block is placed on the wedge, so it experiences two forces, weight (mg) and the normal force (N) from the wedge. Since the surfaces are frictionless, there is no friction force acting on the block.

The net force acting on the block in the horizontal direction is the product of its mass and acceleration (F = ma). As the block is not slipping on the wedge, the acceleration of the block is equal to the acceleration of the wedge, i.e., 10 m/s^2.

Therefore, the net force acting on the block in the horizontal direction is given by:

F = ma = 1 kg x 10 m/s^2 = 10 N

The normal force acting on the block is perpendicular to the surface of the wedge. Therefore, the work done by the normal force on the block is zero. This is because the displacement of the block is perpendicular to the direction of the normal force.

**Work done by the normal force in the ground frame:**
The normal force is an internal force between the block and the wedge. Therefore, the work done by the normal force on the block is zero in any frame of reference, including the ground frame.

Hence, the answer to the problem is zero.

**Conclusion:**
In this problem, we found that the work done by the normal force on the block is zero in any frame of reference, including the ground frame. This is because the normal force is an internal force between the block and the wedge, and the displacement of the block is perpendicular to the direction of the normal force.