Can a telescope detect a blackbody? I'd like to determine the distance at which a telescope with a certain set of parameters (sensitivity, collector area) can detect a blackbody with certain other parameters (temperature, emitting area).

For example, let us suppose we have an infrared telescope with a sensitivity of '1Jy/sr' and a collector area of 10m^2 at a distance D from a 300m^2 perfectly black body at a temperature of 30 Kelvin. 

How do I determine D? What additional information do I need to perform the calculation?
Let's assume a uniform laminar emitter, oriented so that it is at right angles to the line to the detector.

The specific intensity emitted by a blackbody, at the surface of the blackbody is
$$B_{\nu} = \frac{2h\nu^3}{c^2} \frac{1}{\exp[h\nu/kT] -1}\ {\rm Wm}^{-2}{\rm Hz}^{-1},$$
where $B_{\nu}$ is the "Planck function".

Using the definition of specific intensity we can calculate the flux (energy per unit time, per unit frequency) received at the detector as
$$ f_{\nu} \simeq B_{\nu}A \Delta \Omega \cos \theta\,$$
where $\Delta \Omega$ is the solid angle of the detector subtended at the laminar surface, $A$ is the area of the laminar surface, and $\theta$ is the angle with respect to a normal from the surface and here, $\cos \theta \simeq 1$.

The solid angle subtended by the detector at the surface is $\Delta \Omega = a/D^2$, where $a$ is the area of the detector, but we can also define a solid angle $\Delta \omega = A/D^2$, which is the solid angle of the surface at the detector. Thus
$$\frac{f_{\nu}}{\Delta \omega} =  B_{\nu}a  $$

If the detector has a fixed sensitivity per unit solid angle then unless the object is unresolved, then it does not matter how far away it is; the value of $D$ does not come into it. It is a well-known fact in astronomy that resolved, extended objects have an observed surface brightness (flux density per unit solid angle) that is independent of distance.