A potential difference of 6.3 V across a resistor makes a current of 0.9 A flow.

Ohm's law: at a steady temperature, $V = IR$. So $I = \dfrac{V}{R}$ and $R = \dfrac{V}{I}$, with $R$ in ohms ($1\ \Omega = 1\ \text{V/A}$). The V-I graph of such a conductor is a straight line through the origin; its slope is $R$.