Answer
$6.68\;\rm m$
Work Step by Step
To find $d$, we need to find the speed of the block at the top of the incline.
We can choose the system to be the block+ the spring+ the Earth, but there is work done by friction force on our system.
Thus,
$$K_i+U_{ig}+U_{is}+W_{f}=K_f+U_{fg}+U_{fs}$$
We know that the block is released from rest. We chose the initial height of the system to be the origin at which $y=0$.
We also know that the elastic potential energy is zero when the block no longer touches the spring.
Thus,
$$0+0+U_{is}+W_{f}=K_f+U_{fg}+0\tag1$$
The work done by the friction force is given by $W_f=f_k\Delta s\cos180^\circ$ where $\Delta s$ is given from the geometry of the right triangle that is represented by the tilted incline surface.
$\sin45^\circ=\dfrac{h}{\Delta s}$ Thus, $\Delta s=\dfrac{h}{\sin45^\circ }$
$$W_f=-\dfrac{hf_k}{\sin45^\circ}$$
Plugging into (1);
$$ \frac{1}{2}kx_i^2-\dfrac{hf_k}{\sin45^\circ}=\frac{1}{2}mv^2+mgh $$
Noting that $y_i=0$ and $y_f=h$
Solving for $v$;
$$v=\sqrt{\dfrac{\frac{1}{2}kx_i^2-\dfrac{hf_k}{\sin45^\circ}-mgh}{\frac{1}{2}m}}\tag 2$$
Now we need to find the friction force.
$$f_k=\mu_kF_n$$
Now we need to find the normal force exerted on the block while it is moving up the inclined surface.
We chose the perpendicular line to the inclined surface to be the positive $y'$-direction. We know that the net force exerted on the block in this direction is zero.
Thus,
$$F_n=mg\cos45^\circ$$
Therefore,
$$f_k=\mu_kF_n=\mu_kmg\cos45^\circ$$
Plugging into (2);
$$v=\sqrt{\dfrac{\frac{1}{2}kx_i^2-\dfrac{h\mu_kmg\cos45^\circ }{\sin45^\circ}-mgh}{\frac{1}{2}m}} $$
Plugging the known;
$$v=\sqrt{\dfrac{\frac{1}{2}(1000)(0.15)^2-\dfrac{(2)(0.2)(0.2)(9.8)\cos45^\circ}{\sin45^\circ}-(0.2)(9.8)(2)}{\frac{1}{2}(0.2)}}$$
$$v=\bf8.09\;\rm m/s$$
Now we have a new stage of motion. A 200-g block is released at speed of 8.09 m/s at an angle of 45$^\circ$. We need to find the maximum horizontal distance traveled by this block.
First, we need to find the time interval of this trip.
We can find the time it takes from the releasing point to the moment it touches the ground again by finding the time it takes to reach the maximum height and then double it.
$$v_{fy} =v_{iy} +a_yt_1$$
$$0 =v\sin45^\circ-gt_1$$
Thus,
$$t_1=\dfrac{v\sin45^\circ}{g}$$
Hence, the time of the whole trip is given by
$$t=2t_1=2\dfrac{v\sin45^\circ}{g}=2\times\dfrac{8.09\sin45^\circ}{9.8}$$
$$t=\bf 1.167\;\rm s$$
Thus,
$$d=v_xt=v\cos45^\circ \times 1.167$$
$$d=8.09\cos45^\circ \times 1.167$$
$$d=\color{red}{\bf 6.68}\;\rm m$$
