Since the line lies on the given plane, therefore the point $\left(a,2d,c\right)$ on the line also lies on the plane

$\Rightarrow a^2-2bd+c^2=0$

Also, the line will be perpendicular to the normal vector of the plane

$\Rightarrow a^2-b^2+c^2=0$

Hence, $b^2=2bd\Rightarrow \frac{b}{d}=2\Rightarrow \frac{b}{4d}=\frac{1}{2}$