Answer: 314

Solution:

Follow the four edges. Sides $AB$ and $DC$ lie on the lines $y = \frac{x}{3}$ and $y = 3x,$ which map to the lines $v = \frac{u}{\sqrt{3}}$ and $v = \sqrt{3}\,u$ — rays from the origin at angles $30^\circ$ and $60^\circ.$ Sides $AD$ and $BC$ lie on $x + y = 1200$ and $x + y = 2400,$ which map to arcs of the circles $u^2 + v^2 = 1200$ and $u^2 + v^2 = 2400.$

So the image is the part of the annulus between radii $\sqrt{1200}$ and $\sqrt{2400}$ lying between the $30^\circ$ and $60^\circ$ rays, one twelfth of the full annulus: $$\begin{aligned} k &= \frac{30}{360}\,\pi\,(2400 - 1200) \\ &= 100\pi \approx 314.16. \end{aligned}$$

The greatest integer not exceeding $k$ is $314.$