In cylindrical polar-coordinates ($\rho$, $\phi$, $z$) with $\mathbf{n}$ along the $z-$axis, we would get
$$\mathbf{A} = \frac{\rho}{2} \hat{\phi}$$
and along the circle $K$ of radius $\rho=R$,
$$d\mathbf{r} = R d\phi$$
Then
$$\oint_K d\mathbf{r} \cdot \mathbf{A} = \frac{R^2}{2} \int_{0}^{2\pi} d\phi = \pi R^2$$
Now using a vector identity
$$\nabla \times \mathbf{A} = \frac{1}{2} \nabla \times ( \mathbf{n} \times \mathbf{r} ) = \frac{1}{2} \Big[ \hat{z} (\nabla \cdot \mathbf{r}) - (\hat{z} \cdot \nabla) \mathbf{r} \Big] = \hat{z}$$
So that
$$\int_{S} d\mathbf{S} \cdot \nabla \times \mathbf{A} = \pi R^2$$
where the surface $S$ is a cylinder oriented along the $z-$axis with one (open) end on the circle $K$ and the other end closed at an arbitrary distance $z$ from the circle.
In the form presented, $A$ is the vector potential for a uniform magnetic field along the $z-$direction (and in this case of unit strength). I think this is called the radial gauge.