Prashant,
Polymath is right that if the box and man are in equilibrium then the man is pulling the rope with a force of 45g and this does not show the true weight of the man on the machine. So the only possibility is to be in a dynamic equilibrium, so the man and the box will be rising with accesleration a.
Let us analze both cases separately. You can separate out the Man and the box as separate bodies.
On the man the forces are T(Tension) upwards by the rope that he is puliling down. A normal reaction N upwards and of course his weigth 60g downwards.
On the box the forces are T(Tension) upwards by the rope connected to the box to the pulley, the normal reaction N downwards and its weight 30g downwards.
For case 1 if the box is in static equilibrium then the forces should balance and cancel out hence we get the equations:
On Man: T+N=60g
On Box: N+30g=T
Solving we get N=15g, T=45g
since N=15g (weight shown on weighing machine is=15kg) it is not the true weight of the man.
Now suppose the man is pulling the rope with a continuous force irrespective of his motion, with a force T (not equal to 45g since now we assume the man is moving with the box upwards). Assuming the man and box are moving with an acceleration 'a' upwards, then the motion equations become:
On Man: T+N-60g=60a
On Box: T-N-30g=30a
Since we want N to be 60g (to show the true weight of the man), we can solve for T and 'a' to get:
T = 60a and a=3g
Thus T=180g
So if the man pulls the rope with a force 180g continuously then the machine will show his true weight.