Answer
(a) $3200\space N/m$
(b) $1.4\space \mu J$
Work Step by Step
For a spring, $F=kx$
For a solid cylinder, $F=(\frac{YA}{L_{0}})\Delta L$
After comparing the above two equations we can find $x$ is analogous to $\Delta L$ & $k$ is analogous to $(\frac{YA}{L_{0}})$.
(a) $k=\frac{YA}{L_{0}}=\frac{Y(\pi r^{2})}{L_{0}}$ ; Let's plug known values into this equation.
$k=\frac{(3.1\times10^{6}N/m^{2})\pi(0.091\times10^{-2}m)^{2}}{2.5\times10^{-2}m}\approx3200\space Nm$
(b) $x=\frac{F}{k}=\frac{0.03\space N}{3200\space N/m}\approx9.4\times10^{-5}m$
The work done (W) by the variable force is equal to the area under the F-versus-x curve.
So, $W=\frac{1}{2}Fx=\frac{1}{2}\times0.03\space N\times 9.4\times10^{-5}\space m=1.4\times10^{-6}J=1.4\space \mu J$
