Answer: 572

Small Hint:

Write $z=x+iy$; the first condition gives a square, and the second gives two circle inequalities

Big Hint:

Subtract from the $40$-by-$40$ square the union of two semicircles, accounting for their lens-shaped overlap

Solution:

Write $z=x+iy.$ The condition on $\frac{z}{40}$ gives $0\leq x\leq40$ and $0\leq y\leq40.$ Since $$\frac{40}{\overline z}=\frac{40x}{x^2+y^2}+i\frac{40y}{x^2+y^2},$$ the other condition requires $x^2+y^2\geq40x$ and $x^2+y^2\geq40y.$ Thus, within the square, we remove two semicircles of radius $20.$

Their overlap is the lens formed by two radius-$20$ circles whose centers are $20\sqrt2$ apart. Its area is $200\pi-400.$ Hence the removed union has area $400\pi-(200\pi-400)$, or $200\pi+400.$ Let $K$ denote the area of $A.$ Then $$\begin{aligned}K&=1600-(200\pi+400)\\&=1200-200\pi\\&\approx571.68.\end{aligned}$$ The nearest integer is $572.$