Resistance.

Here's a bit of a cheat solution, which doesn't require any calculus. Consider the case in which $a= b$ . The resistor will be a cylinder, and it's resistance will obviously not be infinite. This rules out any candidate solution with $(a-b)$ in the denominator. We know that the cross-sectional area in this case is $\pi a^2$ and not $2 \pi a$ . This leaves only $\frac{\rho l}{\pi a b}$ as the answer.

But in case we didn't want to cheat, here's a better solution:

The radius as a function of position (call it $x$ ) is:

$\large{r = a + (b-a)\frac{x}{l}}$

The cross-sectional area as a function of position is:

$\large{A = \pi r^2 = \pi \Big(a + (b-a)\frac{x}{l}\Big)^2}$

The differential resistance is:

$\large{dR = \frac{\rho dx}{A} = \frac{\rho dx}{\pi \Big(a + (b-a)\frac{x}{l}\Big)^2}}$

The total resistance is:

$\large{R = \int_0^l \frac{\rho dx}{A} = \int_0^l \frac{\rho dx}{\pi \Big(a + (b-a)\frac{x}{l}\Big)^2}}$

Use a variable transformation:

$\large{u = a + (b-a)\frac{x}{l} \\ du = \frac{b-a}{l} dx \\ dx = \frac{l du}{b-a}}$

$\large{R = \frac{\rho l}{\pi (b-a)} \int_a^b \frac{du}{u^2} = \frac{\rho l}{\pi (a-b)} \frac{1}{u} \Bigg|_a^b \\ = \frac{\rho l}{\pi (a-b)} \Big(\frac{1}{b} - \frac{1}{a}\Big) = \frac{\rho l}{\pi ab (a-b)} (a-b) = \frac{\rho l}{\pi ab} }$