The current density in a cylindrical wire of radius $$r = 4.0$$ mm is $$1.0 \times 10^6$$ A$$^2$$ m$$^2$$. The current through the outer portion of the wire between radial distances $$\frac{r}{2}$$ and $$r$$ is $$x\pi$$ A; where $$x$$ is ______

Since the current density in the cylindrical wire is uniform, $$J = 1.0 \times 10^6$$ A/m$$^2$$, and the radius of the wire is $$r = 4.0$$ mm $$= 4.0 \times 10^{-3}$$ m, we need to find the current through the outer portion between $$\frac{r}{2}$$ and $$r$$.

For a uniform current density, the current through an annular region between radii $$r_1$$ and $$r_2$$ is given by

$$I = J \times A = J \times \pi(r_2^2 - r_1^2)$$

Here $$r_1 = \frac{r}{2} = 2.0 \times 10^{-3}$$ m and $$r_2 = r = 4.0 \times 10^{-3}$$ m. Using these values, the area of the outer portion becomes

$$A = \pi(r^2 - (r/2)^2) = \pi\left(r^2 - \frac{r^2}{4}\right) = \pi \times \frac{3r^2}{4}$$

Substituting $$r = 4.0 \times 10^{-3}$$ m gives

$$A = \frac{3\pi}{4} \times (4.0 \times 10^{-3})^2 = \frac{3\pi}{4} \times 16 \times 10^{-6} = 12\pi \times 10^{-6} \text{ m}^2$$

Therefore, the current through this region is

$$I = J \times A = 1.0 \times 10^6 \times 12\pi \times 10^{-6} = 12\pi \text{ A}$$

Comparing this with $$x\pi$$ A yields $$x = 12$$.

The answer is $$\boxed{12}$$.