Key idea

In the usual experimental setup for measuring an unknown resistance $R$ using an ammeter and a voltmeter, if the voltmeter is connected across the combination of ammeter + resistor, then the voltmeter reads the total potential drop $V = I(R + r_A),$ where $r_A$ is the internal resistance of the ammeter.

Hence, the slope of the plotted $V$-$I$ graph is $\frac{V}{I} = R + r_A.$

Using the graph

From the straight-line $V$-$I$ graph, the slope is read as $\text{slope} = 3000\ \Omega.$

So, $R + r_A = 3000\ \Omega.$

Ammeter resistance from circuit data

From the given circuit, the ammeter has internal resistance $r_A = 500\ \Omega.$

Calculate the unknown resistance

Therefore, $R = 3000 - 500 = 2500\ \Omega.$

Final answer

$\boxed{2500\ \Omega}$