CHEOPS (also 1, 2, 3) is a space telescope recently put in low Earth orbit to look for exoplanets around other stars. It uses a cooled detector array and so uses a heat shield to avoid thermal radiation from the Sun as well as the Earth directly below.
Thermal radiation from the Earth is a significant problem for spacecraft thermal management in low Earth orbit, so it's important to know how much thermal radiation Earth emits into space.
If we use an average emissivity $\epsilon$ of 0.8 and assume a temperature of 0 °C, then using
$$P= \sigma \epsilon T^4$$
where the Stefan-Boltzmann constant $\sigma$ is 5.67 $\times 10^{-8}$ W m-2 K-4 we can estimate it to be about 250 W/m2.
But what is it really, and how does it vary (on average) by the type of terrain under a satellite and time of day?
From this answer to For a given 48 hour observing period, what fraction of the celestial sphere is available to CHEOPS? (originally from the CHEOPS Red Book ESA/SRE(2013)7 November 2013