The extension of a uniform spring due to its own weight $Mg$ is given by: $$\Delta l_{self} = \frac{Mg}{2k}$$ Given:

• Mass of spring $M = 50.0 \, \text{g}$
• $\Delta l_{self} = 2.00 \, \text{cm}$

This implies that the effect of the spring’s own weight is equivalent to an end load of $Mg/2$.

We want a total extension $\Delta l_{total} = 10.0 \, \text{cm}$. The total extension is the sum of the extension due to self-weight and the extension due to the added load $W$: $$\Delta l_{total} = \Delta l_{self} + \Delta l_{load}$$ $$10.0 \, \text{cm} = 2.00 \, \text{cm} + \Delta l_{load}$$ $$\Delta l_{load} = 8.00 \, \text{cm}$$

We know the relationship between load and extension is linear ($W = k \Delta l$). We can set up a ratio comparing the “effective” self-weight load to the external load:
Extension of 2 cm corresponds to effective load $M/2 = 25 \, \text{g}$. Extension of 8 cm corresponds to external load $m_{load}$. $$\frac{m_{load}}{M/2} = \frac{8 \, \text{cm}}{2 \, \text{cm}} = 4$$ $$m_{load} = 4 \times \frac{M}{2} = 2M$$ $$m_{load} = 2 \times 50.0 \, \text{g} = 100 \, \text{g}$$