A wire of resistance R is bent to form a square ABCD as shown in the figure. The effective resistance between E and C is: (E is mid-point of arm CD)

The total resistance of the wire is R. Since it is bent into a square ABCD, each side has a resistance of R/4.

Point E is the midpoint of side $CD$. This divides the side CD into two segments, DE and EC, each having a resistance of:

$$R_{DE} = R_{EC} = \frac{1}{2} \left( \frac{R}{4} \right) = \frac{R}{8}$$

To find the effective resistance between E and C, we identify two parallel paths:

1. Path 1 (Direct): Only the segment EC.$$R_1 = \frac{R}{8}$$
2. Path 2 (The rest of the square): Segments $$ED \to DA \to AB \to BC$$
3. .$$R_2 = \frac{R}{8} + \frac{R}{4} + \frac{R}{4} + \frac{R}{4} = \frac{R + 2R + 2R + 2R}{8} = \frac{7R}{8}$$

The effective resistance $R_{eq}$ is calculated using the parallel formula:

$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$$

$$\frac{1}{R_{eq}} = \frac{1}{R/8} + \frac{1}{7R/8}$$

$$\frac{1}{R_{eq}} = \frac{8}{R} + \frac{8}{7R}$$

Taking the common denominator $7R$:

$$\frac{1}{R_{eq}} = \frac{56 + 8}{7R} = \frac{64}{7R}$$

Solving for $$R_{eq}$$:

$$\boxed{R_{eq} = \frac{7}{64}R}$$