Answer
$D = 0.64~m$
Work Step by Step
We can consider the torque about a rotation axis at the hinge to find the vertical component $T_y$ of the cable's tension required to maintain equilibrium:
$\tau_{net} = 0$
$(3.0~m)(T_y)-(1.5~m)(500~N) = 0$
$(3.0~m)(T_y)=(1.5~m)(500~N)$
$T_y=\frac{(1.5~m)(500~N)}{3.0~m}$
$T_y = 250~N$
We can find the minimum angle $\theta$ the cable can make with the horizontal:
$sin~\theta = \frac{250~N}{1200~N}$
$\theta = sin^{-1}~(\frac{250~N}{1200~N})$
$\theta = 12.0^{\circ}$
We can find $D$:
$\frac{D}{3.0~m} = tan~\theta$
$D = (3.0~m)~tan~\theta$
$D = (3.0~m)~tan~12.0^{\circ}$
$D = 0.64~m$
