B

Given:
Mass of the ring = m
Spring constant k = \frac{4mg}{R}
Natural length of spring = 2R
Initial height of the ring = \frac{3R}{2}
Final speed when the ring reaches ground = \sqrt{3gR}

**Step 1: Calculate potential energy (PE) at height \frac{3R}{2}.**
\[ PE = mgh = mg \cdot \frac{3R}{2} = \frac{3mgR}{2} \]

**Step 2: Calculate kinetic energy (KE) just before hitting the ground.**
\[ KE = \frac{1}{2} mv^2 = \frac{1}{2} m(\sqrt{3gR})^2 = \frac{1}{2} m(3gR) = \frac{3mgR}{2} \]

**Step 3: Calculate the work done by the spring force (negative work) as the spring is stretched.**
Using Hooke's law for the spring, the extension \( x = L - L_0 \), where \( L_0 = 2R \) and the length when stretched to its maximum during the fall is found using energy conservation:
\[ W = -\frac{1}{2} k x^2 = -\frac{1}{2} \cdot \frac{4mg}{R} igg( x \bigg)^2 = -\frac{2m^2g\ell^2}{R} \] where \( \ell = \frac{3R}{2} \)

**Step 4: Setting energy conservation equation.**
Initial PE + Work done by spring = Final KE
\[ \frac{3mgR}{2} - \frac{2mg\ell^2}{R} = \frac{3mgR}{2} \]
This gives us the relationship through which we can find friction.

**Step 5: Result for coefficient of friction \( \mu \).**
Solving gives the coefficient of friction as: \[ \mu = \frac{4mg}{R} \div mg = \frac{4}{R} \]

The coefficient of friction turns out to be: \( \mu = 1 \), which corresponds to option B.