C

Step 1: Let the initial current be $I$ and the radius be $R$. The magnetic field $B$ at the center of a circular loop is given by the formula:
$$ B = \frac{\mu_0 I}{2R} $$
where $\mu_0$ is the permeability of free space.

Step 2: The magnetic moment $\mu$ for a circular loop is given by:
$$ \mu = I A = I \pi R^2 $$

Step 3: The initial ratio of magnetic field to magnetic moment is:
$$ \frac{B}{\mu} = \frac{\frac{\mu_0 I}{2R}}{I \pi R^2} = \frac{\mu_0}{2\pi R^3} $$

Step 4: Now, if both current and radius are doubled (i.e., $I' = 2I$ and $R' = 2R$), the new magnetic field becomes:
$$ B' = \frac{\mu_0 (2I)}{2(2R)} = \frac{\mu_0 I}{2R} = B $$
and the new magnetic moment becomes:
$$ \mu' = (2I) \pi (2R)^2 = 4 (I \pi R^2) = 4\mu $$

Step 5: The new ratio becomes:
$$ \frac{B'}{\mu'} = \frac{B}{4\mu} = \frac{1}{4} \cdot \frac{\mu_0}{2\pi R^3} $$

Step 6: The final new ratio is therefore:
$$ \frac{\mu_0}{8\pi (2R)^3} = \frac{\mu_0}{8\pi R^3} $$
Thus, the final answer corresponds to Option C: \( \frac{x}{4} \).