Answer
$W=-0.012J$
Work Step by Step
The work needed to be done to ready the pen for writing is $$W=\frac{1}{2}kx_0^2-\frac{1}{2}kx_f^2$$
The spring constant $k=250N/m$, the spring's initial point $x_0=5mm=5\times10^{-3}m$ and final point $x_f=5+6=11mm=1.1\times10^{-2}m$. Therefore, $$W=-0.012J$$
