Answer
a) is correct.
Work Step by Step
According to the work-energy theorem, $$W_{nc}=E_f-E_0$$ $$W_{air}=\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)$$
The skydiver's velocity remains constant, so $\frac{1}{2}m(v_f^2-v_0^2)=0$
We have $h_f-h_0=-325m$ (Since the skydiver falls downward, her final height is less than her initial height):
$$ W_{air}=-mg\Delta h \\ W_{air}=-(92kg)(9.81 m/s^2)(325m)\\
W_{air}=-2.93\times10^5J$$
