This problem involves the equilibrium condition that the sum of torques must equal zero. This would be easier to describe with a diagram, but we can work with it as is.
The first thing we have to determine is all the individual torques acting on this meter stick (which I assume to be held horizontally in equilibrium, since that best fits the description). The expression for torque exerted by a force around a certain pivot is:
τ = Frsinθ
F = magnitude of the force
r = distance from the pivot point to the point where the force is being applied
θ = angle between the F and r
For a case like this, all the r values will be horizontal, and all the forces will be vertical, so the angles between them will always be 90o. Since sin(90o) = 1, we have, for all the torques in this case, a simplified expression:
τ = Fr
So we just have to figure out F and r in each case, as well as the sign associated with that torque (which we will talk about more as we go along).
What forces are being applied to this meter stick around the pivot? First, there is the weight of the 72 g mass hung at the 20 cm mark. The weight of the 72 g mass is: (0.072 kg)(9.8 m/s2) = 0.706 N. This force is applied at the 20 cm mark, for a case where the pivot is at he 40 cm, so the distance from the pivot to the point where this 0.706 N is being applied is 20 cm, or 0.2 m (we are only concerned about the magnitude of this length, not any kind of direction; we will deal with the direction of this torque by other means, coming up).
From these numbers, we get:
τ72 = Fr = (0.706 N)(0.2 m) = 0.14 N·m
This is the magnitude of the torque, but what about its direction? The kind of rotation we are talking about here is basically like motion along a one-dimensional line; there are only two ways it can go. In this case, those two ways are clockwise and counterclockwise, which we can assign signs. By convention, we usually call torques that cause counterclockwise rotations positive, and those that cause clockwise rotations negative.
With the 72 g mass hanging at the 20 cm mark for a pivot at the 40 cm mark, you may be able to see with a diagram or a prop that -- acting on its own -- this torque would cause the ruler to rotate counterclockwise; so this is a POSITIVE torque. Thus, we have in the the end:
τ72 = (+)0.14 N·m
Now we just have to do the same thing for all the other torques, to eventually build an equation which will allow soling for m.
The unknown mass m is at 10 cm. So its r value is 30 cm = 0.3 m, and its weight is m(9.8 m/s2). This torque -- acting on its own -- would, like the last one, also make the ruler rotate counterclockwise; so it, too, is a positive torque, which we can represent as:
τm = (+)m(9.8)(0.3) = (+)2.94m N·m
That is the end of the hanging masses... but there is one more torque being applied to this meter stick. Any extended massive object, like this ruler, has a center of mass, defined to be the position where -- for the purposes of using Newton's laws) we can consider any external force translating the object to be applied. For the center of mass of the ruler, that force is gravity. We consider the weight of the ruler, therefore, to be exerted at the center of mass. Since the stick is described as "uniform," this means the center of mass is at its center -- the 50-cm mark.
So gravity is considered to be applied to the meter stick at the 50-cm position, which is offset from the pivot point. So the weight of the stick itself exerts another torque on the meter stick. The r-value for this torque is 10 cm = 0.1 m, and the magnitude of the weight is 0.090 kg * 9.8 (m/s2) = 0.882 N·m. This torque alone would cause the meter stick to rotate clockwise about its pivot, so this torque would be considered negative. Thus:
τw = -(0.882 m)(0.1 m) = -0.088 N·m
Now, the condition for equilibrium is that all these torques must add up to zero. This gives:
τnet = τ72 + τm + τw = 0
0.14 + 2.94m - 0.088 = 0
This equation will let you solve for m.
The problem I saw here is that, with the given information, the 72 g mass hung at 20 cm already exerts more torque than the weight of the stick other side of the pivot (because, though the mass is a bit smaller, it is hung twice as far away). So to balance the stick, the extra mass m would have to be hung on the other side of the pivot from what is indicated in the problem. Solving the above equation gives a negative value for the mass, meaning the weight of the mass would have to pull upward to balance the torques.
It would also work out if the ruler were actually a TWO-meter stick, because then its own weight would exert greater torque around a pivot at the 40-cm mark.
Could you please check the problem wording again to make sure it is accurate? If it is, then I would say there is a typo, or it is meant to point out its own implausibility.