The circuit looks like this when you consider that the voltmeter has a resistance itself:
simulate this circuit – Schematic created using CircuitLab
Here I assumed the voltmeter has a resistance of 100K Ohms, which if it is 10000 times the R values works out to R=100 Ohms. A moving coil voltmeter might have an R value of 100K Ohms, whereas a modern FET based digital voltmeter more likely in the 10M Ohms range.
The voltage across the battery is fixed at 9 V if we ignore internal resistance and we have a simple voltage divider across the battery.
The voltage across Node A-B is
NodeA-B = 9V * 100000/100300
= 8.97 V
The answer is very close to 9 V in this instance with these values, and if you consider the voltmeter as 'ideal' ( R tends to infinity) the NodeA-B voltage will tend ever closer to 9V.
You should read up on the voltage divider rule to understand the formulae involved.