Maya walks completely around the boundary of a square whose sides are each $6$ km long. From any point on her path she can see exactly $2$ km horizontally in all directions. What is the area of the region consisting of all points Maya can see during her walk, expressed in square kilometers and rounded to the nearest whole number?

1. If the square has side $s$ and the viewing radius is $r$, the total visible area is $8sr+(\pi-4)r^2$ (an outer expansion by $r$ plus an inner strip of width $r$).
2. With $s=6, r=2$: area $=8(6)(2)+(\pi-4)(4)=96+4\pi-16=80+4\pi\approx92.57$.
3. Rounded to the nearest whole number, the answer is $\boxed{93}$.