$^{-1}$. The reading on the scale is:

There are two forces acting on the man: his weight (due to gravity), which is always downwards, and the normal force from the scale, which is perpendicular to the scale and upwards. When the lift is stationary (not moving), these two forces balance each other out and the reading on the scale is simply the man's weight, which is:

Weight = mass x gravitational acceleration = 70 kg x 9.8 m s$^{-2}$ = 686 N

However, when the lift is moving upwards with a uniform speed of 10 m s$^{-1}$, there is an additional force acting on the man, which is the force due to the lift's acceleration. Since the lift is moving upwards, this force is also upwards, and its magnitude is:

Force due to lift's acceleration = mass x lift's acceleration = 70 kg x 10 m s$^{-2}$ = 700 N

The net force acting on the man is the vector sum of his weight and the force due to the lift's acceleration. Since these two forces act in opposite directions, their vector sum is simply the difference between their magnitudes:

Net force = Weight - Force due to lift's acceleration = 686 N - 700 N = -14 N

Note that the net force is negative, which means that the direction of the net force is downwards. This makes sense, since the man would feel slightly lighter (i.e. experience a smaller normal force) when the lift is moving upwards compared to when it is stationary, due to the lift's upward acceleration partially cancelling out the force due to gravity.

The reading on the scale is equal to the magnitude of the normal force from the scale, which is equal and opposite to the magnitude of the net force acting on the man:

Reading on scale = magnitude of normal force = |Net force| = |-14 N| = 14 N

Therefore, the reading on the scale is 14 N, which is slightly less than the man's weight when the lift is stationary.