Perhaps the original request is a little confuse, but I think I get what Beron wants. Now, the area of a sphere (no planet is a perfect sphere, but close enough) is 4πR^2. This means that if the total area of the planet is 515 000 000 km2, then its radius is the square root of the Area/4π, which is approximately equal to 6402 km (I guess it's ok to be approximate, since the planet is not a perfect sphere to begin with), which is in fact just a little bigger than Earth's. A radius of 6402 gives a circumference (2πR) of approximately 40225 km. So, let's say that the Equator of this planet is 40225 km long. Then, assuming you are using a rectangular projection in which the scale is constant at the equator (which is most of them, I guess, or at least most of those that are most commonly used), that is the width of the map at the equator. If you use an A4 size in landscape mode, that means that 297 mm = 40225 km, thus 1mm =~ 135,4 km, 1cm =~ 1354 km. 1cm^2 =~ 1833316 km^2. If you count by pixels instead of mm just make the appropriate proportion with the width of 40225km.
Unless I made some errors with the calculation, but you get the idea anyway.