Hi
You make a good observation. If there is no voltage across the 3k, it can effectively be removed from the circuit. In an ideal world, I would like to do mesh analysis with the 3k in situ, but this makes it too hard (!).
Therefore, ignoring the 3k, you have 2 strings of 10k (6k + 4k). These 2 are in parallel (so = 5k), so you have a load on the 2k resistor of 5k.
Your total load on the 30V source is 2k + 5k = 7k, so your current is 30/7k = 4.28mA.
Alternatively...
Current going into the 2k resistor is Io. Voltage across the bridge, lets call Vb.
Current into each leg of the bridge is Vb/10k
Using nodal analysis: (30-Vb)/2k = Vb/10k + Vb/10k = Vb/5k
Therefore, 30/2k = Vb(1/5k + 1/2k)
From this you can work out Vb (= 21.428V), then either divide (30 - 21.428V) by 2k to get Io, or divide 21.428V by 5k. Both ways gives you 4.28mA.
Thanks for the refresher - years since I did that!