#### Answer

(a) $I = 6.9~kg~m^2$
(b) $I = 4.1~kg~m^2$

#### Work Step by Step

(a) We can find the moment of inertia of the door when the axis is on its hinges. Let $L$ be the width of the door.
$I = \frac{1}{3}ML^2$
$I = \frac{1}{3}(25~kg)(0.91~m)^2$
$I = 6.9~kg~m^2$
(b) The moment of inertia about an axis through the midpoint of the door's width is $I_{cm} = \frac{1}{12}ML^2$. If the vertical axis is 15 cm from one edge of the door, we can find the distance $x$ from the midpoint to to the axis of rotation.
$x = \frac{0.91~m}{2}-0.15~m$
$x = 0.305~m$
We can use the parallel axis theorem to find the moment of inertia about the axis that is 15 cm from the edge of the door.
$I = I_{cm}+Mx^2$
$I = \frac{1}{12}ML^2+Mx^2$
$I = \frac{1}{12}(25~kg)(0.91~m)^2+(25~kg)(0.305~m)^2$
$I = 4.1~kg~m^2$