Answer
$138\;\rm m/s$
Work Step by Step
The skydiver will reach his terminal speed when his gravitational force (downward) and the drag force (upward) are equal and then the net force exerted on him is zero.
$$\sum F_y=D-F_G=0$$
Thus,
$$D=F_G$$
and we know that the drag force is given by
$$D=\frac{1}{2}C\rho A v^2$$
where $C$ is the e drag coefficient, $\rho$ is the air density, $A$ is the area of the object that faces the air, $v$ is the falling object speed.
and the gravitational force of the skydiver is given by $$F_G=mg$$
Hence,
$$\frac{1}{2}C\rho A v_T^2=mg$$
Therefore,
$$v_T=\sqrt{\dfrac{2mg}{C\rho A}}\tag 1$$
Now we need to find the surface area of the skydiver that faces the air. We know that it falls feet first.
The whole volume of the skydiver is modeled as a rectangular box. And since his feet dive first, his area is given by
$$A=20\times10^{-2}\cdot 40\times10^{-2}=\bf \color{blue}{0.08}\;\rm m^2$$
Plugging the known into (1) and the area from the previous blue result.
$$v_T=\sqrt{\dfrac{2\cdot 75\cdot 9.8}{0.8\cdot 1.2\cdot 0.08 }} =\color{red}{\bf 138}\;\rm m/s$$
