What's the value of the supply voltage in this circuit?

By sensible inspection of the circuit you can say that there must be 2mA through the bottom 3k resistor to produce the output voltage of 6V. That leaves 1mA through the top resistor as 3mA - 2mA = 1mA. 1mA through the top resistor drops 3V from the 6V output voltage leaving 3V as Vs.

Here's a bit more complicated example I created last night. See if you can follow how the mesh equations were derived.

It doesn't matter which way around you draw the mesh current arrows. They can be either way and can even be in different directions to each other. It is OK to step around the loop in either direction to create the mesh equations, the important thing is that you get the signs correct. Voltage rises are positive, voltage drops are negative in sign

Ignore the down marker. Some people become unpleasant when they think you shouldn't be helped so much and they see help being provided.

Well, using Thevenin and Norton we have the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

For $text{R}_text{th}$, we get:

^{simulate this circuit}

So:

$$text{R}_text{th}=frac{text{R}_1text{R}_2}{text{R}_1+text{R}_2}tag1$$

And for $text{I}_text{th}$, we get:

^{simulate this circuit}

So:

$$text{I}_text{th}=text{I}_{text{o}_1}+text{I}_{text{o}_2}=frac{text{V}_text{in}}{text{R}_1}+text{I}_2tag2$$

So, we also know:

$$text{V}_text{th}=text{I}_text{th}cdottext{R}_text{th}=left(frac{text{V}_text{in}}{text{R}_1}+text{I}_2right)cdotfrac{text{R}_1text{R}_2}{text{R}_1+text{R}_2}tag3$$

Using the given values, gives:

$$6=left(frac{text{V}_text{in}}{3000}+3cdot10^{-3}right)cdotfrac{3000cdot3000}{3000+3000}spaceLongleftrightarrowspacetext{V}_text{in}=3spacetext{V}tag4$$