One thing that may help you a lot in the future is to realize that you can label one (and only one) of your "nodes" (read as "wires") as being exactly zero volts (\$0\:\text{V}\$.) So your schematic can be re-drawn like this:

simulate this circuit – Schematic created using CircuitLab
On the left, I simply called the bottom wire "ground" and broke the wire connections, which aren't really needed. (They are all labeled as "ground" so they connect together, whether or not I draw a wire to show that fact.) But it is otherwise the same circuit.
On the right, I just slightly moved things around and then drew a box around an ideal voltage supply with a voltage divider. The only reason for moving things around was to make the voltage divider aspect just a little more obvious, in case it wasn't beforehand. But again, this is otherwise the exact same circuit.
Most textbooks and classwork quickly gets to the point of teaching about voltage dividers. For example, in "The Art of Electronics," 3rd edition, they discuss it on page 7. Which is right after they introduce the idea of micro- and milli- and such prefixes. In other words, very early on. So I'm pretty sure you know what one is. In the same book on page 12, the authors are already telling you about the Thevenin equivalent for it, as well. So I think you should be able to work out the following:

simulate this circuit
Here, I've replaced the voltage divider circuit on the left side with the new Thevenin equivalent on the right side. Please note that this does NOT look like what you drew as your second, lower drawing.
Now, you have a current sink and since this circuit has all its two-lead components in series with each other, you know what the circuit's current is. It's given to you on a silver platter.
You should be easily able to start at the Thevenin voltage on the left side of the right-side circuit shown above and compute the voltage drop across the Thevenin resistance to work out the voltage that must be at node A. Do you think you can do that?