Answer:
[tex]104\; {\rm g}[/tex], assuming that the meter stick is uniform with the center of mass precisely at the [tex]50\; {\rm cm}[/tex] mark.
Explanation:
Refer to the diagram attached. The meter stick could be considered as a lever. The string at the [tex]30\; {\rm cm}[/tex] mark would then act as the fulcrum of this lever.
The [tex]m_{A} = 20\; {\rm g}[/tex] mass at the [tex]80\; {\rm cm}[/tex] mark is at a distance of [tex]r_{A} = 50\; {\rm cm}[/tex] to the right of the fulcrum at [tex]30\; {\rm cm}[/tex].
The weight of the [tex]80\; {\rm g}[/tex] meter stick acts like a weight of [tex]m_{B} = 80\; {\rm g}[/tex] attached to the center of mass of this meter stick. Under the assumptions, this center of mass of this meter stick would be at the [tex]50\; {\rm cm}[/tex] mark, which is [tex]r_{B} = 20\; {\rm cm}[/tex] to the right of the fulcrum at [tex]30\; {\rm cm}[/tex].
Let [tex]m_{C}[/tex] be the mass attached to the meter stick at the [tex]5\; {\rm cm}[/tex] mark. This mass would be at a distance of [tex]r_{C} = 25\; {\rm cm}[/tex] to the left of the fulcrum at [tex]30\; {\rm cm}[/tex].
At equilibrium:
[tex]\begin{aligned} & m_{C}\, r_{C} && (\text{mass on the left of fulcrum})\\ &= m_{A}\, r_{A} + m_{B} \, r_{B} && (\text{mass on the right of fulcrum})\end{aligned}[/tex].
Solve for [tex]m_{C}[/tex], the unknown mass attached to the meter stick at the [tex]5\; {\rm cm}[/tex] mark:
[tex]\begin{aligned}m_{C} &= \frac{m_{A}\, r_{A} + m_{B}\, r_{B}}{r_{C}} \\ &= \frac{20\; {\rm g} \times 50\; {\rm cm} + 80\; {\rm g} \times 20\; {\rm cm}}{25\; {\rm cm}} \\ &= 104\; {\rm g}\end{aligned}[/tex].
