Answer
$k=303N/m$
Work Step by Step
Before the trigger is pulled, we choose the height $h_0=0$ and the spring is compressed by $y_0=9.1\times10^{-2}m$.
When the pellet rises to its maximum height $h_f=6.1m$, the spring is completely released and unstrained, so $y_f=0$
At both points, the pellet has zero speed, so $KE=0$. There is also no rotational kinetic energy. Since total energy is conserved, $$\frac{1}{2}ky_0^2=mgh_f$$ $$k=\frac{2mgh_f}{y_0^2}$$
The pellet's weight $mg=(2.1\times10^{-2}kg)\times(9.8m/s^2)=0.206N$. Therefore, $$k=303N/m$$
