Answer
$13\;\rm m/s^2$
Work Step by Step
First of all, we need to draw a force diagram of the shuttle.
When the suttle is released, the elastic cord will pull the shuttle up while pulling it to the right.
This means that the normal force is acting downward, as you see below.
We analyzed the initial tension force into its components as you see in the second figure below.
The net force exerted on the shuttle vertically is zero since it has no motion in this direction.
$$\sum F_y=T\sin45^\circ -F_n-mg=ma_y=m(0)=0$$
$$T\sin45^\circ -F_n-mg=0$$
Solving force the normal force;
$$F_n=T\sin45^\circ -mg \tag 1$$
The net force exerted on the shuttle horizontally is given by
$$\sum F_x=T\cos45^\circ-f_k=ma_x$$
Recall that $f_k=\mu_kF_n$
$$T\cos45^\circ-\mu_kF_n=ma_x$$
Plugging $F_n$ from (1);
$$T\cos45^\circ-\mu_k (T\sin45^\circ -mg )=ma_x$$
Solving for $a_x$;
$$a_x=\dfrac{T\cos45^\circ-\mu_k (T\sin45^\circ -mg )}{m} $$
Plugging the known;
$$a_x=\dfrac{20\cos45^\circ-0.6(20\sin45^\circ -[0.8\cdot 9.8])}{0.8} $$
$$a_x=\color{red}{\bf 13}\;\rm m/s^2$$
