Answer
$ 2.25\times 10^5\;\rm \Omega^{-1}\cdot m^{-1}$
Work Step by Step
We know, from Ohm's law, that
$$\Delta V=IR$$
So if this material is an ohmic one, the graph $\Delta V-I$ will give us a straight line that has a slope of $R$. And if the material is not an ohmic one, then the graph will not be linear.
We plugged the 4 points into the $\Delta V-I$ graph, as we see below.
And the best-fit line is already a straight line and it started from the origin as expected, so the material is an ohmic one.
As we can see the slope is 0.4 which is $R$
$${\rm Slope}=R=\bf 0.4\;\rm \Omega$$
Now we need to find the conductivity of this material. Recalling that the resistance is given by
$$R=\dfrac{\rho L}{A}$$
where $\rho$ is the resistivity which is given by $\rho=1/\sigma$ where $\sigma$ is the conductivity.
$$R=\dfrac{ L}{\sigma A}$$
Thus,
$$\sigma=\dfrac{ L}{R A}$$
Plug the known;
$$\sigma=\dfrac{ (45\times 10^{-3})}{(0.4)(0.5\times 1.0\times 10^{-6}) }$$
$$\sigma=\color{red}{\bf 2.25\times 10^5}\;\rm \Omega^{-1}\cdot m^{-1}$$
