The center of gravity of a meter stick is at the 50 cm mark. If the pivot support is located at the 50 cm mark and a 50 g mass is attached to the 20 cm mark, how much mass would have to be attached to the 65 cm mark to place the meter stick in rotational equilibrium?

This has to do with balancing torques about the pivot point, so the first weight times its distance from the pivot point must be balanced by the second weight times its distance from the pivot point. Draw a sketch and it will help you see what's going on. Post your attenopt and we'll check it for you.