Kirchhoff's laws in an open circuit
Here's what you know and have already worked out:







simulate this circuit &ndash; Schematic created using CircuitLab

The only thing I did was to add a grounding reference point (in the lower left corner.) It's not necessary. But who knows? It may help you "see" better. By the way, you get to assign \$0\:\text{V}\$ to any one node of a circuit, your choice. But only one. This simplifies and improves communication -- voltages are always measured between two nodes but if you assign one node to be zero, then all the other node voltages can be specified with reference to that special node. It helps a little.

Since there is no voltage drop across \$R_5\$, it follows that the node between \$R_5\$ and \$V_2\$ is also at \$0\:\text{V}\$. It then follows that \$V_b=+4\:\text{V}\$ (with reference to our newly located ground reference position.) I think you should be able to see that much.

Next, since the current in \$R_3\$ is \$5\:\text{mA}\$ and directed downwards, it follows that the top node of \$R_3\$ is \$5\:\text{mA}\cdot 200\:\Omega=+1\:\text{V}\$. (The bottom node of \$R_3\$ is, by definition, at \$0\:\text{V}\$.)

Finally, since there is also no voltage drop across \$R_4\$, it follows that \$V_a=+1\:\text{V}\$ (with reference to our ground reference position.)

So, you know that \$V_a=+1\:\text{V}\$ and \$V_b=+4\:\text{V}\$. You should be able to work out either \$V_{ab}=V_a-V_b\$ or \$V_{ba}=V_b-V_a\$ on your own from that information.