Fair enough, but you're not going to like the result. I was too conservative late last night with just how much light the object would be receiving and consequently reflecting. Here's all the numbers worked out for you. Enjoy!

Given
b = L/(4*pi*d^2)
where b = brightness (flux)
L = luminosity (in watts)
d = distance (in meters)
Solar luminosity = 3.839 × 10^26 watts
200 AU distance = 2.99196 × 10^13 meters
That works out to a solar flux of 0.034 watts/m^2 at 200 AUs distance, so that is the solar flux this hypothetical object would receive. Given a surface area of 6.1419 × 10^16 m^2 and divide that by two (since the object would receive light from the sun over half its surface at any given time), then multiply by the flux, that works out to a total of 1.044123 × 10^15 watts that the object receives. Given TrES-2b's best fit albedo of 0.04% (0.0004) it will reflect 4.176492 x 10^11 watts of light. Now plug that number back into the original formula again with the same 200 AU distance to calculate the flux received from this object by an observer at earth. The result is 3.712703 x 10^-17 watt/m^2

given that mx = -2.5log(Fx/Fox)

where mx = magnitude difference between object x and reference object
Fx = flux of object x
Fox = flux of reference object

Using the sun as a reference object, and given the sun's flux of 1370 watts/m^2 at earth, plugging in the flux for the object shows the difference between the sun's magnitude and the object's magnitude being 48.9175759. The sun has an apparent magnitude at earth of -26.74, so the magnitude that is 48.9175759 magnitudes dimmer is magnitude 22.1775759, even brighter than I originally approximated last night. It also happens to be about the same magnitude as asteroids that are detected by amateurs using half meter instruments in 5 minute exposures.