Under the influence of an applied horizontal force, a block of mass...

As the block of mass $m$ slips on the rough horizontal table, the kinetic frictional force exerted by the table on the block is $f = \mu mg$.

According to Newton's third law of motion, the block exerts an equal and opposite frictional force of magnitude $f = \mu mg$ on the table in the horizontal direction.

Since the table remains stationary on the floor, its horizontal acceleration is zero, meaning the net horizontal force acting on the table must be zero.

The only horizontal forces acting on the table are the friction from the block and the static friction from the floor. Therefore, the total frictional force exerted by the floor on the legs of the table must perfectly balance the frictional force exerted by the block.

Thus, the magnitude of the frictional force exerted by the floor is $\mu mg$.