A metal wire of length L, area of cross-section A and Young's modulus Y behaves as a spring of spring constant k.

For an external force F, let the wire extension be x.

$$\begin{gathered} \mathrm{stress}=\frac{\mathrm{F}}{\mathrm{A}}; \\ \mathrm{strain}=\frac{\mathrm{x}}{\mathrm{L}}; \end{gathered}$$

$$\begin{gathered} \mathrm{Y}=\frac{{\displaystyle \frac{\mathrm{F}}{\mathrm{A}}}}{{\displaystyle \frac{\mathrm{x}}{\mathrm{L}}}}; \\ \mathrm{Y}=\frac{\mathrm{FL}}{\mathrm{xA}}; \\ F=\frac{YA}{L}x \\ \mathrm{Equivalent} \mathrm{forece} \mathrm{constant} = k=\frac{F}{x}=\frac{YA}{L} \end{gathered}$$

Hence the correct answer is $k=\frac{YA}{L}.$