EM radiation has a relativistic mass (see for instance, Does a photon exert a gravitational pull?), and therefore exerts a gravitational pull.
Intuitively it makes sense to include EM radiation itself in the galactic mass used to calculate rotation curves, but I've never actually seen that done before...
So: if we were to sum up all the electromagnetic radiation present in a galaxy, what fraction of the dark matter would it account for?
Answer
I found it surprisingly hard to find an authoritative statement of the density of the CMB. According to this article it's about $5 \times 10^{-34}\mathrm{g\ cm}^{-3}$, and since the critical density is somewhere in the range $10^{-30}$ to $10^{-29}\mathrm{g\ cm}^{-3}$ photons don't make a significant contribution.
Photons wouldn't be dark of course. If there were enough photons to make a significant contribution to the mass/energy of the universe we'd see them, just as we can see the CMB.
Response to comment: oops yes, I didn't read your question properly - sorry!
Anyhow, my comment that photons aren't dark matter still applies, but it's easy to make an estimate of the gravitational contribution of the EM radiation in e.g. the Solar System. The Sun converts about $4 \times 10^9$ kg of matter to energy every second. Since it weighs about $2 \times 10^{30}$ kg every second it loses about $2 \times 10^{-19}$% of it's mass every second.
If you're prepared to assume the photon density in the Solar System is dominated by the Sun's output (which seems plausible) and take the size of the Solar System to be Neptune's orbit, i.e. $1.5 \times 10^4$ light seconds then the mass/energy of photons in the Solar System is $3 \times 10^{-15}$% of the Sun's mass. So it's utterly insignificant.
The reason photons make a much lower contribution to the Solar System than to the universe as a whole is because mass is much more concentrated in the Solar System than in the universe as a whole.
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