Answer
$ {\bf 1.79\times 10^8}\;\rm A/m^2$
Work Step by Step
We know, according to Ohm's law, that
$$\Delta V=IR$$
where $R=\rho L/A$
$$\Delta V=I\dfrac{\rho L}{A}$$
where $I/A$ is the current density $J$
Thus,
$$\Delta V=\left(J\rho \right)L$$
Replacing $L$ by $s$ to match the given graph,
$$\Delta V=\left(J\rho \right)s$$
This is a formula of a straight line where $\left(J\rho\right)$ is the slope of the given graph.
$${\rm Slope}=J\rho_{\rm tungsten} $$
Hence,
$$J=\dfrac{{\rm Slope}}{\rho_{tungsten}}$$
Plug the known from the figure and Table 30.3;
$$J=\dfrac{\dfrac{3-0}{0.3-0}}{5.6\times 10^{-8}}$$
$$J=\color{red}{\bf 1.79\times 10^8}\;\rm A/m^2$$
