Answer
(c) B,C,A
Work Step by Step
$ \frac{1}{R_{eqA}} = \frac{1}{R} + \frac{1}{R}+ \frac{1}{2R} $
$ R_{eqA} = \frac{2R}{5}$
$ \frac{1}{R_{eqB}} = \frac{1}{2R} +\frac{1}{2R} $
$ R_{eqB} = R$
$ \frac{1}{R_{eqC}} =\frac{1}{R}+\frac{1}{3R}$
$R_{eqC} = \frac{3R}{4}$
Since $ R \gt \frac{3R}{4} \gt \frac{2R}{5} $
Therefore $ R_{eqB} \gt R_{eqC} \gt R_{eqA}$
