At which setting do you expect the resistance to be higher?

In order to avoid confusion between electrical potential and its unit, the volt, I will use ξ for the former and V for the latter.

The electrical potential (voltage) is clearly the same for both:

ξ = 120 V

You just need Ohm’s Law:

ξ = IR

and the fact that electrical power used in maintaining current, I,

across a voltage drop, ξ, is the product of the two:

P = ξI

Then, just combine them to eliminate the non-constant, I:

I = ξ/R

P = ξ²/R

Showing that, at constant voltage, the higher power occurs with the lower resistance.

At each setting, ξ² = 14400 V²

Then

at P = 950 W, . . . R = ξ²/P = 15.2 Ω

at P = 1450 W, . . R = ξ²/P = 9.93 Ω

The resistance is higher when the setting is at 950 W.

Resistance at 950 W:

P= (V^2) / R

R=15.2 ohms

Resistance at 1450 W:

P= (V^2) / R

R=9.93 ohms