The correct option is:
(a)
(B) it will remain at rest, and the force of friction will be 13.7 N.
(b)
(A) it will accelerate under a net force of about 1.9 N.
The magnitude of static force of friction on block is:
f_s = μ_sN
Here, N is the normal force on the block.
The block rest on a level surface, therefore the normal force should be equal to the weight of the block, that is, N = mg .
Substitute the same in equationf_s = μ_sN.
f_s = μ_sN.
Here, m is the mass of the block, and g is the acceleration due to gravity.



Round off to three significant figures,
f_s = 15.7 N
Similarly, the magnitude of force of kinetic friction is given as:



Round off to three significant figures,
f_k = 11.8 N
(a)
It is important to note that the applied force F = 13.7 N is greater than the kinetic force of friction f_k = 11.8 N, and is capable of moving the object. However the object moves when the applied force has taken over the rest state of the object appearing due to static force of friction.
Therefore, to move the object, the applied force is initially larger than force of static friction, and once the motion has started, the block can be moved if the magnitude of applied force is greater than kinetic force of friction. From this, one can deduce that the applied force F = 13.7 N is insufficient in moving the object against static friction f_s = 15.7 N, and cannot move the object.
Therefore, the object remains at rest and experiences a static friction of magnitude equal to the magnitude of applied force, that is,13.7 N.
Thus, (B) is the correct option, and the rest are ruled out.

Therefore, the block will accelerate with a net force of 1.9 N.
Thus, (A) is the correct option, and the rest are ruled out.