You underestimate the total resistance with your approach as you do not consider current flow in radial direction (which contributes to resistance, too). This is easier to see if you take extreme cases - attach two small disks of radius r (the electrodes) to a cylinder with radius R>>r and height h~r. If you increase R, your calculated resistance would drop with 1/R^2, but the real resistance will approach a finite value.
I did some two-dimensional numerical simulation, but the borders are messy to consider.
For b=0.96, I get ~2.3 ρ/a as resistance, where your formula gives 1.24. I think my simulation overestimates the resistance a bit, so the real value is somewhere in between.
For b=0.52, I get ~0.38 ρ/a as resistance, where your formula gives 0.367. Here, the radial resistance is not significant.
Fixing a=ρ=1 now, as they are just prefactors anyway:
For (b-1)<<1, it is possible to calculate an interesting lower bound of the resistance as following:
Replace the disk (electrode) by a half-sphere. This replaces conducting material by an ideal conductor and clearly reduces the resistance.
Replace the half-ball by a half-ball around the electrode and connect the second electrode to the outer boundary. Use this as half the full resistance. This adds conducting materials where the original problem has none and moves the original symmetry area towards the electrode, so it again underestimates the resistance.
This modified problem has a spherical symmetry, so we can solve it with a one-dimensional integral. Let r be the radius of the electrode, ##r^2+b^2=1##.
$$R > 2 \int_r^1 \frac{1}{2\pi r'^2} dr' = \frac{1}{\pi} (\frac{1}{r}-1) = \frac{1}{\pi}(\frac{1}{\sqrt{1-b^2}}-1)$$
Compare this with your formula at b=0.99:
R > 1.93
R with your formula: 1.68
It gets worse with larger b, as the radial flow gets more important:
b=0.9999:
R > 22.2
R with your formula: 3.15
