How much watts per square meter does Phobos receive from Mars? Being so close to Mars, Phobos should receive a non-negligible amount of irradiance from Mars. Be it from reflected sunlight, Mars' emitted infrared radiation, or other factors. I have no idea how to calculate those however.
 A: There are a lot of variables (distance of Mars from sun, place of Phobos in orbit etc.) but we can do a back of the envelope calculation using data from Wikipedia here, here and here
The solar radiation received by Mars at the top of its atmosphere varies between $492$ and $715$ Watts per square metre. For the purposes of an upper limit estimate let's take the maximum of $715$ Watts per square metre.
The Bond albedo (average albedo across all wavelengths) of Mars is $0.25$. Its visual geometric albedo is smaller, but again for an upper limit let's take the higher figure. So the solar radiation reflected by Mars is about $179$ Watts per square metre.
The radius of Phobos's orbit is about $2.7$ times the radius of Mars. Using the inverse square law, the radiation intensity at Phobos will be smaller than that at Mars by a factor of $2.7^{-2} \approx 0.14$. So the maximum intensity at the surface of Phobos of solar radiation reflected from Mars is about $715 \times 0.25 \times 0.14 \approx 25$ Watts per square metre.
For comparison the maximum solar irradiance on the Earth at sea level on a clear day is about $1,000$ Watts per square metre.
A: Since you have "no idea" how to address this, it's good to start with the fundamentals and make some order-of-magnitude estimates and bounds.
First is the sun's contribution. The sun's output is:
$$ L = 3.8\times 10 ^{26}\,{\rm W}$$
The irradiance at Mars is:
$$ I =\frac 1 {4\pi R^2} L = 571\,{\rm W/m^2}$$
where $R$ is the radius of Mars's orbit (roughly).
In the 1st estimate, we'll treat Phobos as so close to Mars that the planet occupies $2\pi$ str.
The upper limit of reflected radiation occurs if Mars is 100% reflective, in which case it acts like a mirror and Phobos sees:
$$ I_{reflected} \le I = 571\,{\rm W/m^2}$$
(It is possible that diffuse, but perfect, reflection will include a factor of 1/2, that is TBD)
The other limit is if Mars is a blackbody. In this case, we can use the energy absorbed:
$$ E = \pi r^2 \times I = 2 \times 10^{16}\,{\rm W} $$
which is then re-emitted from its entire surface area, for a flux density:
$$ E/A = \frac E{4\pi r^2} = \frac I 4 = 142\,{\rm W/m^2}$$
which, according to $\sigma T^4$, equates to a temperature of:
$$ T = \big(\frac{E/A}{\sigma}\big)^{\frac 1 4} = 224 K $$
Now, according to our approximation, Phobos sees a wall of 224 K blackbody. Integrated over $2\pi$ steradian, that will emit $I_{BB} = 142 {\rm W/m}^2$.
From that you can (roughly) estimate the total flux as:
$$ I = \epsilon I_{BB} + (1-\epsilon)I_{reflected} $$
where $\epsilon$ is the emissivity.
From here, you can relax the assumption that Mars is so close that it occupies $2\pi$ str. While a full answer requires integrating over a curved surface with different distance factors, including Lambertian scattering $1/\cos{\theta}$ factors, the 1st approximation to a correction factor can use the ratio of the orbit size ($R_p$) to Mars:
$$ f = \big(\frac r {R_p}\big)^2 \approx 1/7 $$
which is probably too small: Mars is not a point source. One could use the solid angle subtended by Mars (relative to $2\pi$):
$$ f = \frac 1 {2\pi}\times 4\pi\sin^2{\tan^{-1}\frac r {R_p}} \approx 1/4$$
which is an upper bound.
The main take aways are: (1) that reflected light directly illuminates Phobos (on the day side, it is of course zero at night), while blackbody radiation is reemitted in all directions. (2) All the solar radiation that falls on Mars is emitted in some form.
