The resistance of a conductor at 15°C is 16 $$\Omega$$ and at 100°C is 20 $$\Omega$$. What will be the temperature coefficient of resistance of the conductor?

We are given that the resistance of the conductor is $$R_1 = 16\;\Omega$$ at the initial temperature $$T_1 = 15^{\circ}\text{C}$$ and becomes $$R_2 = 20\;\Omega$$ at the higher temperature $$T_2 = 100^{\circ}\text{C}$$.

For a metallic conductor, the variation of resistance with temperature is described by the linear relation

$$R = R_0\left[1 + \alpha\,(T - T_0)\right],$$

where

$$R_0$$ = resistance at the reference temperature $$T_0,$$

$$R$$ = resistance at temperature $$T,$$

$$\alpha$$ = temperature coefficient of resistance (per °C).

Applying this formula to our data, we write

$$R_2 = R_1\left[1 + \alpha\,(T_2 - T_1)\right].$$

Substituting the known values, we get

$$20 = 16\left[1 + \alpha\,(100 - 15)\right].$$

First evaluate the temperature difference:

$$100 - 15 = 85^{\circ}\text{C}.$$

So the equation becomes

$$20 = 16\left[1 + 85\,\alpha\right].$$

Now divide both sides by 16 to isolate the bracketed term:

$$\frac{20}{16} = 1 + 85\,\alpha.$$

Simplify the left-hand fraction:

$$\frac{20}{16} = 1.25.$$

Thus we have

$$1.25 = 1 + 85\,\alpha.$$

Subtract 1 from both sides:

$$1.25 - 1 = 85\,\alpha.$$

$$0.25 = 85\,\alpha.$$

Finally, divide by 85 to solve for $$\alpha$$:

$$\alpha = \frac{0.25}{85}.$$

Perform the division step by step:

$$\alpha = \frac{25}{8500} = \frac{1}{340} \;\text{per}\;^{\circ}\text{C}.$$

Numerically,

$$\alpha \approx 0.00294\;^{\circ}\text{C}^{-1} \approx 0.003\;^{\circ}\text{C}^{-1}.$$

This matches the third option given in the list.

Hence, the correct answer is Option C.