Measuring Resistance of a Wire With an ADC
Please, don't use an LM324 if you want to do precision measurements.  

Your opamp has a gain of 5, but you're not using that: Your output is the inverting input, where you have the same signal as the non-inverting, so that's gain x 1.  

The best choice would be an instrumentation amplifier, where you connect the cable's ends to the two inputs. Use a series resistor to ground to create an offset, because InAmps can't go to the rails (at least the 3-opamp types can't). You can use that resistor as a sense resistor for the current source:  



\$V_{IN}\$ sets the current of the current source: 100 mA/V. Suppose the cable's resistance is 5 &Omega;, then the InAmp will see a 500 mV difference on its input. A gain of 10 (gain resistor isn't shown; CircuitLab doesn't have a symbol for InAmps) will give you 5 V out, or 1 V/&Omega;. By changing \$V_{IN}\$ you can change the total gain. Note that Q1 may need a heatsink, especially if Vcc is rather high.  

 

If you expect high resistances you can make a resistor divider with 1 precision resistor to Vref, and one to ground:



The voltage across the cable will be

\$ V_{CABLE} = \dfrac{R_{CABLE}}{R_{CABLE} + 2 R} V_{REF} \$

but if \$R_{CABLE}\$ &lt;&lt; \$2 R\$ the voltage may be too low for an accurate measurement. A low value for \$R\$ helps, but will draw much current.

The MCP6N11 has Rail-to-Rail output and exists in different types for different gains, among which one for a gain of minimum 100.

edit
markrages comments that we don't need an InAmp, and he's right. Here's the solution with a differential amplifier using an opamp:  



The gain is determined by R1 through R4, and if R1 = R3 and R2 = R4 will be   

\$ G = \dfrac{R2}{R1}\$

An InAmp will give you more precision though, and it won't cost you an arm and a leg, so why not?