Solution:

To find the voltage across the current source, we can find the equivalent resistance of the entire circuit. Taking this a step at a time...

First, we can combine the 3k and 7k series resistors (3k + 7k = 10k), and combine the resulting series equvalent resistor value with the 10k resistor in the original circuit (10k || 10k = 5k). You have now combined three resistors (3k, 7k, 10k) into a single equivalent resistance (5k):

At this point, there are no more parallel/series reductions we can make. Now, we can make a difficult circuit more workable by applying a Y-delta transform to the three resistors at the edge of the circuit:

In this example, we are transforming R1, R2, and R3 in the "Y" circuit to an equivalent "delta" circuit of Ra, Rb, and Rc values. Since R1, R2, and R3 are all equal, Ra, Rb, and Rc will also all be equal, and we only have to calculate one of the values:

Ra = [5k*5k + 5k*5k + 5k*5k] / 5k = 75k/5k = 15k

Now, we have two pairs of resistors that we can combine in parallel: 5k || 15k = 3.75k. Redrawing the circuit, again, we've run out of series/parallel resistors to simplify. So, again, let's apply a Y-delta transform. This time, there's a bit more math, b/c the resistor values aren't equal:

So, using the the node numbers above, Ra sits between nodes 2 and 3, Rb site between nodes 1 and 3, and Rc sits between nodes 1 and 2. Ra = 3.75k, Rb = 15k, and Rc = 3.75k.

Using equations in Y-Δ transform - Wikipedia:

R1 = (Rb*Rc) / (Ra + Rb + Rc) = (15k * 3.75k) / (3.75k + 15k + 3.75k) = 56.25k / 22.5k = 2.5k

R2 = (Ra*Rc) / (Ra + Rb + Rc) = (3.75k * 3.75k) / (3.75k + 15k + 3.75k) = 14.0625k / 22.5k = 0.625k

R3 = (Ra*Rb) / (Ra + Rb + Rc) = (3.75k * 15k) / (3.75k + 15k + 3.75k) = 56.25k / 22.5k = 2.5k

Once we've transformed the "delta" circuit to a "Y" circuit, we now have several series/parallel simplifications that can be made. First, combine 5k and .625k into an equivalent 5.625k resistor value. Then, similarly, 5k and 2.5k into 7.25k value. These two series equivalent values are now in parallel, and can be combined: 5.625k || 7.25k = (5.625k * 7.5k) / (5.625k + 7.25k) = 42.1875k / 13.125k = 3.2143k.

At last, all we have left is series resistors: 1k + 2.5k + 3.2143k = 6.7143k.

So, to find the voltage across the current source, we use the current value and the resistor value to find the voltage across the equivalent resistor: 1mA * 6.7143k = 6.7143 V.

I converted 3 delta networks to y networks. On the surface this seems a more complicated solution, however because all resistors are the same value, only one calculation is needed for the 3 delta conversions to y. (5x5/(5+5+5)).