If you want higher accuracy, start with an 8.66Ω resistor, measure it's actual resistance accurately and calculate the required parallel resistance to bring it down to 8.5Ω
The benefit of this method is that the tolerance of the much higher value parallel resistor will have relatively little effect on the total resistance.
As an example, suppose the actual resistance was 8.635Ω. This requires a 543.7Ω resistor in series. Let's assume we have a 549Ω 1% resistor (this is an E96 value). It could be between about 543Ω and 555Ω. The worst value would be 555Ω, and even at this value the total resistance would be 8.5027Ω, a 0.03% error, a difference likely to be swamped by lead resistance and the resistance of the solder joint!
Let's assume that all we had was a 5% 560Ω resistor. At it's maximum likely value of 588Ω, the combined resistance would be 8.5100Ω, still only a tad over 0.1% high.