Answer
$19.6\;\rm N/m$
Work Step by Step
We need to find the initial speed of the abll $v_0$.
We know the initial height of the ball, so we can find the time it takes to reach the ground since we know the horizontal distance as well.
We assume that air resistance is negligible, so the system [the ball+the Earth] is conserved.
Using the kinematic formula of dispalcement.
$$x=x_0+v_{ix}t+\frac{1}{2}a_xt^2$$
The origin is the release point.
$$x=0+v_{0}\cos\theta\; t+\frac{1}{2}(0)t^2$$
$$x= v_{0}(\cos\theta) t $$
Plugging the known;
$$5= v_{0}(\cos30^\circ) t $$
Thus,
$$t=\dfrac{5}{v_{0}(\cos30^\circ)}\tag 1$$
$$y=y_0+v_{iy}t+\frac{1}{2}a_yt^2$$
$$0=1.5+v_{0}(\sin\theta) t-\frac{1}{2}gt^2$$
Plugging from (1);
$$0=1.5+ \color{red}{\bf\not} v_{0}\sin30^\circ \left[\dfrac{5}{ \color{red}{\bf\not} v_{0}(\cos30^\circ)}\right]-\frac{1}{2}g\left[\dfrac{5}{v_{0}(\cos30^\circ)}\right]^2$$
$$0=1.5+5\tan30^\circ-\dfrac{25g}{2v_0^2(\cos30^\circ)^2}$$
$$ 1.5+5\tan30^\circ=\dfrac{(25\times 9.8)}{2v_0^2(\cos30^\circ)^2}$$
$$2v_0^2(\cos30^\circ)^2 \left[1.5+5\tan30^\circ\right]=245$$
Thus,
$$v_0 =\sqrt{\dfrac{245}{2(\cos30^\circ)^2 \left[1.5+5\tan30^\circ\right]}}=\bf 6.102\;\rm m/s$$
Now we need to apply the conservation of fenrgy law from the moment at which the spring was compressed to the moment the ball is released.
In this case, we chose the origin to be the initial position of the ball when the spring is compressed. Hence, we need to find the final height of the ball since it moves up the inclined surface for a distance of 20 cm.
$$\sin30^\circ =\dfrac{h}{\Delta S}$$
Thus,
$$h=y_f=\Delta S\sin30^\circ\tag 2$$
Therefore,
$$E_i=E_f$$
$$K_i+U_{si}+U_{gi}=K_f+U_{sf}+U_{gf}$$
The initial velocity of the ball when the spring was compressed is zero, so the initial kinetic energy is zero. And the initial gravitational potential energy is zero since we chose this point to be our origin at which $y=0$. The final elastic potential energy of the spring is zero since it is not compressed nor stretched when the ball is fired out.
$$0+U_{si}+0=K_f+0+U_{gf}$$
$$\frac{1}{2}k(\Delta S)^2=\frac{1}{2}mv_0^2+mgh$$
Plugging from (2);
$$\frac{1}{2}k(\Delta S)^2=\frac{1}{2}mv_0^2+mg\Delta S\sin30^\circ$$
Solving for $K$;
$$ k= \dfrac{mv_0^2+2mg\Delta S\sin30^\circ}{(\Delta S)^2}$$
Plugging the known;
$$ k= \dfrac{(0.02\times 6.102^2)+(2\times 0.02\times 9.8\times 0.2\sin30^\circ)}{0.2^2}$$
$$k=\color{red}{\bf 19.6}\;\rm N/m$$