The correct answer is:
$$\oint {\overrightarrow {\text{A}} \cdot \overrightarrow {{\text{d}}l} } = \mathop{{\int!!!!!\int}\mkern-21mu \bigcirc} {\left( {\nabla \times \overrightarrow {\text{A}} } \right) \cdot \overrightarrow {{\text{ds}}} } $$
This is known as Stokes’ theorem. It states that the line integral of a vector field around a closed loop is equal to the surface integral of the curl of the vector field over any surface bounded by the loop.
The curl of a vector field is a measure of how much the vector field rotates around a point. It is defined as the cross product of the gradient and the vector field.
The surface integral of a vector field over a surface is the sum of the dot products of the vector field and the area element over the surface.
In this case, the vector field is $\overrightarrow {\text{A}} \left( {\overrightarrow {\text{r}} } \right)$ and the surface is the closed surface bounded by the loop. The curl of the vector field is $\nabla \times \overrightarrow {\text{A}} $.
The line integral of $\overrightarrow {\text{A}} \cdot \overrightarrow {{\text{d}}l} $ around the loop is equal to the surface integral of $\left( {\nabla \times \overrightarrow {\text{A}} } \right) \cdot \overrightarrow {{\text{ds}}} $ over the closed surface bounded by the loop.
The other options are incorrect because they do not take into account the closed surface bounded by the loop.