Funny Wire

The resistance is proportional to $R \propto \int_0^L \frac{dx}{(D(x))^2},$ where $x$ runs along the length of the wire and $D(x)$ is the diameter of the wire at that point.

Taking $L = 1$ , we have $D_A(x) = 2$ and $D_B(x) = 1 + 2x$ . So $R_A \propto \int_0^1 \frac{dx}{2^2} = \frac 1 4; \\ R_B \propto \int_0^1 \frac{dx}{(1+2x)^2} = \left[-\frac 1 {2\cdot (1+2x)}\right]_0^1 \\ = \left(-\frac 1 6\right) -\left(- \frac 1 2\right) = \frac 13$ where the proportionality constants are equal. Therefore

$\frac{R_B}{R_A} = \frac{1/3}{1/4} = \frac 4 3,$ so that $R_B = \boxed{4.00}\ \Omega$ .