1. Resistance of one copper wire

For a wire, $R = \rho \frac{L}{A}$ where $\rho$ is resistivity, $L$ is length, and $A$ is cross-sectional area.

For one copper wire of length $l$ and diameter $d$, $A = \frac{\pi d^2}{4}$ So resistance of one wire is $R_1 = \rho \frac{l}{\pi d^2/4} = \frac{4\rho l}{\pi d^2}$

2. Eight identical wires joined in parallel

If 8 identical resistors each of resistance $R_1$ are connected in parallel, the equivalent resistance is $R = \frac{R_1}{8}$ Thus, $R = \frac{1}{8}\cdot \frac{4\rho l}{\pi d^2} = \frac{\rho l}{2\pi d^2}$

3. Single wire of length $2l$ with same resistance

Let its diameter be $D$. Then its area is $A' = \frac{\pi D^2}{4}$ Its resistance is $R' = \rho \frac{2l}{\pi D^2/4} = \frac{8\rho l}{\pi D^2}$

Given $R' = R$, $\frac{8\rho l}{\pi D^2} = \frac{\rho l}{2\pi d^2}$

Cancel common terms $\rho l/\pi$: $\frac{8}{D^2} = \frac{1}{2d^2}$

So, $16d^2 = D^2$ $D = 4d$

4. Final answer

The required diameter is $\boxed{4d}$ So the blank is 4.