Take downward as the positive $$y$$ direction.
(a) While the woman was in free fall, $$Δy = 144 ft, v_{i} = 0$$, and we take $$a = g = 32.0 ft/s^{2}$$. Thus,
$$\Delta y=v_{i}t+\dfrac{1}{2}at^{2}\rightarrow 144ft=0+(16.0ft/s^{2})t^{2}$$
giving $$t_{fall} = 3.00 s$$. Her velocity just before impact is:
$$v_{f} = v_{i} + gt = 0 + (32.0 ft/s^{2} )(3.00 s) = 96.0 ft/s$$
(b) While crushing the box, $$v_{i} = 96.0 ft/s , v_{f} = 0$$, and
$$Δy = 18.0 in. = 1.50 ft$$. Therefore,
$$a=\dfrac{v_{f}^{2}-v_{i}^{2}}{2(\Delta y)}=\dfrac{0-(96.0ft/s)^{2}}{2(1.50ft)}=-3.07*10^{3}ft/s^{2}$$
or, $$a=3.07*10^{3}ft/s^{2}$$ upwards = $$96.0g$$
(c) Time to crush box:
$$\Delta t=\frac{\Delta y}{\bigtriangledown }=\dfrac{\Delta y}{\dfrac{v_{f}+v_{i}}{2}}=\dfrac{2(1.50ft)}{0+96.0ft/s}$$
or, $$\Delta t=-3.13*10^{-2}s$$