Answer
$36.8\;\rm m/s$
Work Step by Step
He will reach his terminal speed when the net force exerted on him in the $x$-direction became zero.
First, let's draw the force diagram of our system [the skier+skis], as you see below.
Thus,
$$\sum F_x=mg\sin40^\circ -D-f_k=0$$
$$mg\sin40^\circ -D-f_k=0\tag 1$$
We know that the drag force is given by $D=\frac{1}{2}C\rho A v^2$ and for the terminal speed,
$$D=\frac{1}{2}C\rho A v_T^2\tag 2$$
We also know that the kinetic friction force is given by
$$f_k=\mu_k F_n\tag 3$$
We can find the normal force by applying Newton's second law in the $y$-direction.
$$\sum F_y=F_n-mg\cos40^\circ=0 $$
Thus,
$$F_n=mg\cos40^\circ$$
Plug into (3);
$$f_k=\mu_kmg\cos40^\circ \tag 4$$
Plugging (2), and (4) into (1) and then solving for $v_T$;
$$mg\sin40^\circ -\frac{1}{2}C\rho A v_T^2-\mu_kmg\cos40^\circ =0 $$
$$mg\sin40^\circ -\mu_kmg\cos40^\circ = \frac{1}{2}C\rho A v_T^2 $$
$$\dfrac{mg\sin40^\circ -\mu_kmg\cos40^\circ}{ \frac{1}{2}C\rho A} = v_T^2 $$
$$v_T =\sqrt{\dfrac{2mg\left[\sin40^\circ -\mu_k \cos40^\circ\right]}{ C\rho A} }$$
Plugging the known;
$$v_T =\sqrt{\dfrac{2\cdot 80\cdot 9.8\left[\sin40^\circ -0.06 \cos40^\circ\right]}{ 0.8\cdot 1.2\cdot 1.8\cdot 0.4} }$$
$$v_T =\color{red}{\bf 36.8}\;\rm m/s$$
