Answer : 3.375J

The change of length is 5cm =.05m.Since $F=kx$ we have

$45=k(.05)$

$k=\large\frac{4500}{5}$

$\;\;\;=900$ N/m

The work done is

$W=900\int\limits_{0.05}^{.1} xdx$

$\;\;\;\;=\large\frac{900x^2}{2}\bigg]_{0.05}^{0.1}$

$\;\;\;\;=450[.1^2-.05^2]$

$\;\;\;\;=3.375J$