Answer
plane with $y=\dfrac{\sqrt 3x}{3}$
Work Step by Step
In the cylindrical coordinate system, we have $x=r \cos \theta \\ y=r \sin \theta \\z=z$
Conversion of rectangular to cylindrical coordinate system, we have $r^2=x^2+y^2 \\ \tan \theta=\dfrac{y}{x} \\z=z$
Here, the given angle is $\theta=\dfrac{\pi}{6}$ that means that surface will be a plane passing through the origin with an angle $\dfrac{\pi}{6}$ above the $x$-axis in the $xy$ plane.
Now, $\tan (\dfrac{\pi}{6})=\dfrac{y}{x}$
$\implies \dfrac{y}{x}=\dfrac{\sqrt 3}{3}$
Thus, $y=\dfrac{\sqrt 3x}{3}$
Hence, we have an equation of a plane with $y=\dfrac{\sqrt 3x}{3}$.
