Help on this Alevel physics question

Corrections on the diagram -- the resistor R is in series with a voltmeter, not an Ammeter. You can view the voltmeter V1 as another resistor in series with R. R must have very high resistance in order to be compatible. The reading on V2 tells us there is a votage of 3V across R. So the voltage across V1 is 6V. From the ratio of the voltages we can see that the resistance of V1 is twice the combined resistance of R and V2 (In this case the resistance of V2 is not much larger than R, so it has to be considered in the circuit). Therefore resistance of R is qual to that of V1 and V2, which is 10MΩ.

series circuit, 9V power supply, resistor has a resistance of r and 3V going across it calculate r sorry for messy scribbles