Eight copper wires of length $$l$$ and diameter $$d$$ are joined in parallel to form a single composite conductor of resistance $$R$$. If a single copper wire of length $$2l$$ has the same resistance $$R$$ then its diameter will be ______ $$d$$.

The resistance of a wire is given by:

$$R_{\text{wire}} = \dfrac{\rho l}{A} = \dfrac{\rho l}{\pi (d/2)^2} = \dfrac{4\rho l}{\pi d^2}$$

Each wire has length $$l$$ and diameter $$d$$, so its resistance is

$$R_1 = \dfrac{4\rho l}{\pi d^2}$$

When eight such wires are connected in parallel, the equivalent resistance becomes

$$R = \dfrac{R_1}{8} = \dfrac{4\rho l}{8\pi d^2} = \dfrac{\rho l}{2\pi d^2}$$

Now consider a single wire of length $$2l$$ and diameter $$D$$ having the same resistance $$R$$. Its resistance is given by

$$R = \dfrac{4\rho (2l)}{\pi D^2} = \dfrac{8\rho l}{\pi D^2}$$

Equating the two expressions for $$R$$ yields

$$\dfrac{\rho l}{2\pi d^2} = \dfrac{8\rho l}{\pi D^2}$$

which simplifies to

$$\dfrac{1}{2d^2} = \dfrac{8}{D^2}$$

and hence

$$D^2 = 16d^2$$

$$D = 4d$$

Therefore, the diameter of the single wire is $$\boxed{4}d$$.