I have information on the spectral irradiance incoming from the sun at the top of the atmosphere in units of $\rm mW\ m^{-2}\ nm^{-1} $.
The photons hit a hypothetical surface on the earth with 0.3 reflectance (albedo) and are reflected back to the top of the atmosphere, where I need to know the spectral radiance in units of $\rm W\ m^{-2}\ sr^{-1}\ \mu m^{-1} $.
(atmospheric effects are not considered as this all happens in a clear atmospheric window)
Can someone explain how I can do this?
OK, my original answer was totally wrong. Here's my second attempt. Any help is appreciated.
Let us assume that Earth is a Lambertian reflector, which means that the surface scatters light (power) evenly in all directions. There's lots of cosines that will float around, but it works like this:
The latter two cancel out (a property of Lambertian surfaces), and so the outgoing radiance is dependent on the cosine of the angle between the surface normal and the Sun. This means that the radiance you measure is going to depend on that angle. Let us call the direction pointing directly towards the Sun $\varphi = 0$, so that angle runs up to $\varphi = \pi/2$ at the terminator.
We know that incoming solar irradiance is $I_0\approx$1360 W/m$^2$, or whatever number you were provided. Because our surface is Lambertian, we can convert this to outgoing radiance simply by dividing by $\pi$.[1] Let us call this solar radiance $L_0$.
If Earth was an infinite, flat sheet, then the measured $L$ in space would not depend on the inverse square law because the area you view would increase at the same rate the intensity of light decreases. This assumption is still valid as long as we aren't too far from Earth. Thus, all that remains is to average $L$ over the area of Earth's surface that we can see.
Let us denote $R$ as the radius of Earth and $h$ as the height above Earth from which we measure $L$, the radiance. $\varphi$ is the angle from the normal to the Sun, and $\varphi_0$ is the chosen $\varphi$ at which we are measuring $L$. Finally, $\alpha$ is the albedo.
The furthest angle at which our observer will see can be found with simple trig:
$$\varphi_m = \arccos\left(\frac{R}{R+h}\right)$$
This gives us a spherical cap of area $A$,
$$A = 2\pi R^2(1-\cos\varphi_m) = 2\pi R^2\left(1-\frac{R}{R+h}\right)$$
Now, remember that the incoming irradiance depends on $\cos\varphi$ as
$I = I_0\cos\varphi$
So we need to integrate over the spherical cap like
$$L = \frac{I_0}{\pi A}\int_0^{2\pi}\int_0^{\varphi_m}\cos\varphi R^2\sin\varphi\,d\varphi d\theta$$
right? Well it's not so simple. This only works of we choose $\varphi_0 = 0$. If we want to measure elsewhere, we have to rotate our coordinate system. The cosine effect depends on the coordinate system relative to $\varphi = 0$, but our integration takes place with respect to $\varphi = \varphi_0$. I posted on the Math Stack the required formula to make the switch (after some struggle!).
$$\cos\varphi = \frac{1}{2}\sin(2\varphi_0)(1-\cos\theta')\sin(\varphi_0+\varphi') + \left[\cos\theta' + \cos^2(\varphi_0)(1-\cos\theta')\right]\cos(\varphi_0+\varphi')$$
where $\varphi'$ and $\theta'$ are our integration variables w.r.t. $\varphi_0$. I.e. $\varphi' = 0 \leftrightarrow \varphi = \varphi_0$. Now we can perform the integration
$$L = \frac{I_0\alpha}{\pi A}\int_0^{2\pi}\int_0^{\varphi_m}\left[\frac{1}{2}\sin(2\varphi_0)(1-\cos\theta')\sin(\varphi_0+\varphi') + \left[\cos\theta' + \cos^2\varphi_0(1-\cos\theta')\right]\cos(\varphi_0+\varphi')\right] R^2\sin\varphi'\,d\varphi' d\theta'$$
It looks nasty, but MATLAB claims that it simplifies nicely into very simple formula:
$$\begin{aligned} L &= \frac{I_0\alpha R^2}{A}\cos\varphi_0\left[1-\left(\frac{R}{R+h}\right)^2\right]\\ \\ &= \frac{I_0\alpha\cos\varphi_0 (2R + h)}{2\pi(R+h)}\\ \end{aligned}$$
which correctly gives us in the limit $h \ll R$,
$$L \approx \frac{I_0\alpha\cos\varphi_0}{\pi}$$
[1] The surface emits over $2\pi$ steradians, but integrating over that solid angle with the third cosine effect yields a factor of 2, for a total $1/\pi$ factor.
