Explanation

(c) When rod is bent in the form of square, then each side has resistance of
$\frac{1}{4}\Omega.$ As shown $R_{1},\ R_{2}\ $and$\ R_{3}$ are connected in series, so their equivalent resistance
$R^{'} = R_{1} + R_{2} + R_{3}$
$\ \ \ \ \ \ = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}\Omega$
Now, R’ and $R_{4}$ are connected in parallel, so equivalent resistance of the circuit is
$R = \frac{R^{'} \times R_{4}}{R^{'} + R_{4}}$
$\ \ \ \ = \frac{\left( \frac{3}{4} \right)\left( \frac{1}{4} \right)}{\left( \frac{3}{4} \right) + \left( \frac{1}{4} \right)}$
$= \frac{(\frac{3}{16})}{1} = \frac{3}{16}\Omega$