Answer
Choice D.
Work Step by Step
From the graph given, we see that at a temperature of $25^{\circ}C$ and a current of 0.1 A (100 mA), the voltage is about 0.75 V. At $150^{\circ}C$, it is about 0.50 V.
The sensitivity is $\frac{\Delta V}{\Delta T}$.
$\frac{0.50V-0.75 V}{150^{\circ}C -25^{\circ}C }\approx-2\times10^{-3}V/C^{\circ}$
