Answer
$\alpha = 3.0~rad/s^2$
Work Step by Step
If the horizontal pull is negligible, then the only torque on the person is due to the gravitational force.
We can find the initial angular acceleration:
$\tau = I~\alpha$
$\alpha = \frac{\tau}{I}$
$\alpha = \frac{mg~d}{I}$
$\alpha = \frac{(70~kg)(9.8~m/s^2)(0.28~m)}{65~kg~m^2}$
$\alpha = 3.0~rad/s^2$
