proscience - Your idea is very good, and solves the problem of excessive current in the loop (saving battery life, as well). If you choose a resistor with a value of say 100Ω, and if your graphite rod is about 3.2Ω, your total load is 103.2Ω. Using Ohm's law, I=E/R you can calculate the current in the loop as being about 0.088 Amperes or 88 milliamperes. Almost 97% of the voltage from the battery (8.73 volts) will be dropped across the 100Ω resistor and about 0.27 volts across the graphite rod. You can also calculate the power dissapated by the resistor as P=I^2*R, or about 0.77 watts. You should buy a 1 watt rated resistor for the experiment. A multimeter in the 20 volt range will read two places to the right of the decimal. Should be sufficient for this experiment. If your multimeter has a 2 volt range you might consider an even higher value of series resistor, lowering the current and probably reduce the resistor to a 1/4 watt component.
A caution, there is sometimes a confusing understanding between resistivity and what we are measuring, actual resistance. Resistivity deals with the bulk properties of a material to conduct a current. See the attached link so you know how to apply the math.
There are many variants to approaching the problem, and it sounds like you have a pretty good understanding of Ohm's law at this point so you can define a very good experiment yourself. I agree, that the best projects clearly demonstrate that you understand what is happening, and support your hypotheses with a well defined experiment and good measurement and presentation of data.
http://hyperphysics.phy-astr.gsu.edu/hb ... resis.htmlRick Marz