Answer
The magnitude of the torque about the axis due to the door's own weight is $~25.0~N\cdot m$
Work Step by Step
The torque is $\tau = r\times F$, where $r$ is the smallest distance between the force vector and the rotation axis, and $F$ is the perpendicular component of the force vector.
We can assume that the force of gravity pulls down on the door at the door's center of gravity, that is, at the center of the door. Then the smallest distance $r$ between the force vector and the rotation axis is $0.50~m$, which is half the width of the door.
We can find the magnitude of the torque:
$\tau = r\times F$
$\tau = r\times mg$
$\tau = (0.50~m)(50.0~N)$
$\tau = 25.0~N\cdot m$
The magnitude of the torque about the axis due to the door's own weight is $~25.0~N\cdot m$
