Answer
$154.7\;\rm N$
Work Step by Step
First of all, we need to sketch the situation here, as we see in the first figure below.
Then we need to draw the force diagram of the two objects, the box, and the sled, as we see below.
It is obvious now that the only force that prevents the block from sliding down is the static friction force between the box and the sled.
The box will be on the verge of sliding when the net force exerted on it exceeds this static friction force (the maximum static friction force).
$$\sum F_{x,box}=f_{s,box}-mg\sin\theta=m_{box}a_x$$
$$f_{s,box}-m_{box}g\sin\theta=m_{box}a_x $$
Noting that the acceleration of the two objects is the same as long as the box does not slide.
We know that the static friction force is given by $f_s=\mu_sF_n$
$$\mu_sF_{n,box}-m_{box}g\sin\theta=m_{box}a_x \tag 1$$
so we need to find the normal force.
$$\sum F_{y,box}=F_{n,box}-m_{box}g\cos\theta=m_{box}a_y=m_{box}(0)=0$$
Thus;
$$F_{n,box}=m_{box}g\cos\theta\tag 2$$
Note that the normal force exerted on the box is from the sled, and the box will also apply the same force of the sled but downward as we will see below. These two forces are equal in magnitude since they are an action-reaction pair.
Also, note that the static friction force exerted on the box by the sled up the inclined is equal to the static friction force exerted by the box on the sled down the inclined since they are an action-reaction pair.
Now we need to find the tension force by applying
Newton's second law on the sled.
$$\sum F_{x,sled}=T-f_{s,box}-f_k-m_{sled}g\sin\theta=m_{sled}a_x$$
Thus,
$$T=m_{sled}a_x+ f_{s,box}+f_k+m_{sled}g\sin\theta $$
where $f_k=\mu_kF_{n,sled}$, and $f_{s,box}=\mu_s F_{n,box}=\mu_sm_{box}g\cos\theta$
So that
$$T=m_{sled}a_x+\mu_sm_{box}g\cos\theta+\mu_kF_{n,sled}+m_{sled}g\sin\theta\tag 3 $$
Plugging (2) into (1) and then solving for $a_x$;
$$\mu_s\color{red}{\bf\not}m_{box}g\cos\theta-\color{red}{\bf\not}m_{box}g\sin\theta=\color{red}{\bf\not}m_{box}a_x $$
$$a_x=\mu_sg\cos\theta-g\sin\theta$$
$$a_x=g\left[\mu_s\cos\theta- \sin\theta \right]\tag 4$$
Now we need to find $F_{n,sled}$;
$$\sum F_{y,sled}=F_{n,sled}-F_{n,box}-m_{sled}g\cos\theta=m_{sled}a_y=m_{sled}(0)=0$$
Thus,
$$F_{n,sled}=F_{n,box}+m_{sled}g\cos\theta$$
Plugging from (2);
$$F_{n,sled}=m_{box}g\cos\theta+m_{sled}g\cos\theta$$
$$F_{n,sled}=\left[m_{box} +m_{sled}\right]g\cos\theta\tag 5$$
Plugging (4) and (5) into (3);
$$T=m_{sled}g\left[\mu_s\cos\theta- \sin\theta \right]+\mu_sm_{box}g\cos\theta+\\
\mu_kg\cos\theta\left[m_{box} +m_{sled}\right]+m_{sled}g\sin\theta $$
where $\mu_k$ is between ice and wood while $\mu_s$ is between wood and wood.
Plugging the known;
$$T=(20) (9.8)\left[0.5\cos20^\circ- \sin20^\circ \right]+0.5 (10) (9.8)\cos20^\circ+\\
0.06(9.8)\cos20^\circ\left[10+20\right]+(20) (9.8)\sin20^\circ $$
$$T=\color{red}{\bf 154.7}\;\rm N$$
Therefore, at any value greater than that tension force, the box slides.
