Step 1:
Let $CD=x$ then $AB=2CD=2x$.Let $r$ be the radius of the circle inscribed in the quadrilateral ABCD.
Given: Area of quadrilateral ABCD=$18$ and $AB\:is\:||$ to $CD$.
$\Rightarrow \large\frac{1}{2}$$(x+2x).2r=18$
$\Rightarrow\:3xr=18$
$\Rightarrow\:xr=6$-----(1)
$OP=OM=PD=OQ=AM=r$
$\Rightarrow\:PC=x-r\:\: and \:\: MB=2x-r$
Let $\angle PCO=angle\:OCQ=\theta$ then from right-angled $\Delta OPC$
$\tan\theta=\large\frac{OP}{CP}=\frac{r}{x-r}$-----(2)
$CD \parallel AB$
$\therefore \angle PCB=\angle QOM=2\theta$
Step 2:
$\angle CBA=180^{\large\circ}-2\theta$
$\angle OBM=90^{\large\circ}-\theta$
$\Rightarrow\:$ From $\Delta \:OMB,\tan(90^{\large\circ}-\theta)=\large\frac{OM}{MB}=\frac{r}{2x-r}$
From right angled $\Delta OBM$
$\tan\theta=\large\frac{2x-r}{r}$-----(3)
From (2) & (3)
$\large\frac{r}{x-r}=\frac{2x-r}{r}$
$\Rightarrow 2x^2-3xr=0$
$x(2x-3r)=0$
$x=\large\frac{3r}{2}$-------(4)
From (1) & (4) we get,
$xr=6$
$\large\frac{3rr}{2}=$$6$
$r^2 = 4$
$r=2$
Hence (b) is the correct answer.