The resistance of a conductor is 1.08 Ω. To reduce it to 1 Ω, the resistor must be connected with: 

To reduce the resistance value we should connect another resistance in parallel.

When two resistances are connected in parallel combination, the equivalent resistance is given by:
 $\mathrm{R}=\frac{{\mathrm{R}}_1{\mathrm{R}}_2}{{\mathrm{R}}_1+{\mathrm{R}}_2}$

Substituting the given values in the equation, 
$1=\frac{1.08\times {\mathrm{R}}_2}{1.08+{\mathrm{R}}_2}$
$\Rightarrow$$1.08+{\mathrm{R}}_2=1.08{\mathrm{R}}_2$
$\Rightarrow$$1.08=0.08R_2$
$\therefore$ ${\mathrm{R}}_2= \frac{108}{8} = 13.5 \mathrm{\Omega }$ 
 Therefore the conductor with a resistance of 1.08 Ω must be connected with a resistance of 13.5 Ω in parallel combination in order to obtain an equivalent resistance of 1 Ω.