Calculate the resistance and its uncertainty using Ohm's Law, given V = (18 ± 3) V and I = (5 ± 0.1) A.

The potential difference is $V = (18 \pm 3) \text{ V}$.
So, the measured potential difference is $V = 18 \text{ V}$ and its absolute error is $\Delta V = 3 \text{ V}$.

The current is $I = (5 \pm 0.1) \text{ A}$.
So, the measured current is $I = 5 \text{ A}$ and its absolute error is $\Delta I = 0.1 \text{ A}$.

According to Ohm's Law, resistance $R$ is given by:

$$R = \frac{V}{I}$$

When quantities are divided, the maximum fractional error in the result is the sum of the fractional errors of the individual quantities.

$$\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}$$

Fractional error in potential difference:

$$\frac{\Delta V}{V} = \frac{3 \text{ V}}{18 \text{ V}} = \frac{1}{6}$$

Fractional error in current:

$$\frac{\Delta I}{I} = \frac{0.1 \text{ A}}{5 \text{ A}} = \frac{1}{50}$$

$$\frac{\Delta R}{R} = \frac{1}{6} + \frac{1}{50}$$

To add these fractions, find a common denominator, which is 150:

$$\frac{\Delta R}{R} = \frac{25}{150} + \frac{3}{150} = \frac{28}{150} = \frac{14}{75}$$

Percentage error is the fractional error multiplied by $100\%$.

$$\text{Percentage error} = \frac{\Delta R}{R} \times 100\%$$ $$\text{Percentage error} = \frac{14}{75} \times 100\%$$ $$\text{Percentage error} = \frac{14 \times 4}{3}\% = \frac{56}{3}\%$$ $$\text{Percentage error} \approx 18.666...\%$$

Rounding to one decimal place, the percentage error is $18.7\%$.

The percentage error in the measurement of resistance of the conductor is $\boxed{\text{18.7\%}}$.