Answer
$4\; \Omega$.
Work Step by Step
The given values are
$R_1=12 \;\Omega$
$R_2=6 \;\Omega$
Total resistance in parallel circuit:-
$\Rightarrow \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$
Substitute all values.
$\Rightarrow \frac{1}{R}=\frac{1}{12}+\frac{1}{6}$
At the right hand LCD is $12$.
$\Rightarrow \frac{1}{R}=\frac{1}{12}+\frac{2}{12}$
Simplify.
$\Rightarrow \frac{1}{R}=\frac{1+2}{12}$
$\Rightarrow \frac{1}{R}=\frac{3}{12}$
$\Rightarrow \frac{1}{R}=\frac{1}{4}$
$\Rightarrow R=4\; \Omega$.
