1. Use the linear relation for resistance with temperature:

$R_t = R_0(1+\alpha t)$

where $\alpha$ is the temperature coefficient of resistance at $0^\circ \text{C}$.

2. Given:

$R_{15} = 16\,\Omega, \qquad R_{100} = 20\,\Omega$

So,

$16 = R_0(1+15\alpha)$ $20 = R_0(1+100\alpha)$

3. Divide the two equations to eliminate $R_0$:

$\frac{20}{16} = \frac{1+100\alpha}{1+15\alpha}$

$\frac{5}{4} = \frac{1+100\alpha}{1+15\alpha}$

4. Cross-multiply:

$5(1+15\alpha)=4(1+100\alpha)$

$5+75\alpha=4+400\alpha$

$1=325\alpha$

$\alpha=\frac{1}{325}$

$\alpha \approx 0.00308\,^\circ \text{C}^{-1}$

5. Compare with the options:

$\alpha \approx 0.003\,^\circ \text{C}^{-1}$

So the correct option is C.

6. Verification with stored answer:

Stored correct answer = C

This matches our derived answer.