Answer
$(2,\frac{\pi}{2})$.
Work Step by Step
We know that if the rectangular coordinates of the point are $(x,y)$, then $r=\sqrt{x^2+y^2}$ and $\theta=\tan^{-1}{(\frac{y}{x})}$ if the point is not on one of the axes. $r=\sqrt{0^2+2^2}=2$, but here the point is on the y-axis, hence: $\theta=\frac{\pi}{2}$ (because it is on the upper half). The point has polarcoordinates $(2,\frac{\pi}{2})$.
