If I were a beam of light traveling in space, where would I be.?

Raymond is correct. Since time has
stopped for the beam, I would be
all over the Universe at once,

SPACE

Anywhere in a straight line path away from where you started :)

(Unless you passed by a black hole or two, then you might have gotten turned a little)

you would travel at 186,000 miles per second so you would be 186,000 miles away in one second and each second that goes by you would be 186,000 miles further

According to Quantum Theory, you aren't in a specific location until such time as you are measured. Of course this depends a lot on whether you are just one photon or a group of several.

Though light appears to be a 'beam', it consists of particles known as 'photons'.
These photons travel the same way any other electro- magnetic wave does...
In a straight line, unless something else causes a deflection, at 186,220 miles/sec through a vacuum.
You'd be heading in the same direction you were when you left!

REALLY fast!

it depends on your destination I think

I have to go with Aculeus...
but in a strange way.

According to time dilation theories (which were integrated into the theory of Relativity) a 'clock' that would be carried along your moving frame of reference, would appear (to us) to slow down according to the Lorentz transformation equation.

Entanglement-assisted teleportation can be (almost) understood if one accepts that time is stopped for the photon moving at the speed of light:

If a event generates two photons (moving in opposite directions) then these photons will have opposite spins. Let us say that we can force the photon we detect to have a spin of +1, then that forces the other photon to have a spin of -1.
It appears to send information to another photon (already gone) at faster-than-the-speed-of-light. In reality, when the photons left the source (at the time they were still in contact), one was also in contact with our detector (since it is simultaneously at the source and at the detector, in its frame of reference). Therefore, it can "tell" the other photon what spin it must take.

Because light (in vacuum) travels at a constant speed, Special Relativity should be sufficient; therefore we are allowed to use Lorentz's equation.

D'(t) = D(t) / SQRT(1 - v^2)

D(t) is a time interval (Difference in time)
D'(t) is perceived difference in time -- as seen from another frame of reference
1/SQRT(1 - v^2) is the Lorentz factor (a.k.a. "gamma") if v is given as a fraction of c (the speed of light in a vacuum).

Going from A to B as a beam of light moving at some finite speed, we perceive you as moving for a time interval D'(t). We see you move at v=c (=1 in our chosen units of speed).

Within your own frame of reference, you actually experience a true (co-moving) time interval of:

D(t) = D'(t) * SQRT(1 - v^2) = D'(t) * SQRT(1-1) = 0

As far as you are concerned, you are at A at the same time as you are at B (time interval of zero).
Since this will be true for any A and any B along the "beam" of light, then you are simultaneously everywhere along the beam of light (from your point of view).

Time dilation has been shown to be true (e.g., muons generated by cosmic rays do exist much longer than they should if time dilation did not exist).

So, if you are a beam of light travelling in space, make sure you have an anti-beam travelling the other way... it'll be less lonely for the entire 0 seconds of your journey.

Somewhere.

In my mother-in-law's head.