Answer: C. At A, $V/I=2.0/0.40=5.0\ \Omega$ at B, $V/I=5.0/0.80=6.25\ \Omega$ Since the resistance is not constant, the behaviour is non-ohmic. Using $P=IV$ $P_A=0.80\ \mathrm{W}$ and $P_B=4.0\ \mathrm{W}$ so $P_B/P_A=5.0$

A: obtains the power ratio but treats the unequal $V/I$ values as equal, classifying the resistor as ohmic.
B: uses the current ratio, $0.80/0.40=2.0$ as the power ratio and omits the change in potential difference.
D: assumes constant resistance and uses $P\propto I^2$ giving a ratio of $4.0$

Remember: compare $V/I$ to identify ohmic behaviour, and use $P=IV$ for power at each operating point.