1. Given reaction and equilibrium constant

The given reaction is:

$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

with

$K_C = 64$

2. Required reaction

We need the value of $K_C$ for:

$NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)$

3. Relate the required reaction to the given one

First, reverse the given reaction:

$2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$

When a reaction is reversed, the equilibrium constant becomes reciprocal:

$K_C' = \frac{1}{64}$

4. Now divide the entire reaction by 2

Dividing

$2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)$

by 2 gives:

$NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)$

When all stoichiometric coefficients are multiplied by a factor $n$, the equilibrium constant becomes $K^n$.

Here, coefficients are divided by 2, so the new equilibrium constant is:

$K_C'' = \left(\frac{1}{64}\right)^{1/2}$

$K_C'' = \frac{1}{8}$

5. Final answer

$\boxed{\frac{1}{8}}$

So the correct option is B.