The load versus elongation graph for four wires is shown. The...

Using (\quad $\mathrm{Y}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \mathrm{L} / \mathrm{L}}$)
Cross-section area of the wire (\quad $\mathrm{A}=\pi \mathrm{R}^{2}$) where (\quad $\mathrm{R}$) is the radius of the wire (\therefore \quad $\mathrm{Y}=\frac{\mathrm{F}}{\pi \mathrm{R}^{2}} \times \frac{\mathrm{L}}{\Delta \mathrm{L}}$)
(\Longrightarrow $\mathrm{F}=\frac{\pi \mathrm{R}^{2} \mathrm{Y}}{\mathrm{L}} \Delta \mathrm{L}$)
Thus Slope of load vs elongation graph, ($\mathrm{s}=\frac{\pi \mathrm{R}^{2} \mathrm{Y}}{\mathrm{L}}$)
(\Longrightarrow \quad $\mathrm{s} \propto \mathrm{R}^{2}$)
Now from the graph shown, OA has the least slope and hence the least radius. Thus ($\mathrm{OA}$) represents the thinnest wire.