A cylinder of radius $r$ and of thermal conductivity $K_1$ is surrounded by a cylindrical shell of inner radius $r$ and outer radius $2 r$ made of a material of thermal conductivity $K_2$. The effective thermal conductivity of the system is

Consider a length $L$ of the composite cylinder. The system consists of two cylindrical layers in parallel for radial heat flow. The thermal resistance $R$ of a cylindrical shell of inner radius $r_1$, outer radius $r_2$, length $L$, and thermal conductivity $K$ is given by $R = \dfrac{\ln(r_2/r_1)}{2 \pi K L}$.

For the inner cylinder of radius $r$, we consider it as a solid cylinder. However, in the context of effective thermal conductivity for radial heat flow, we treat the cross-sectional area. The total cross-sectional area $A$ of the composite cylinder is $\pi (2r)^2 = 4 \pi r^2$.

The thermal resistance of the inner cylinder of radius $r$ is $R_1 = \dfrac{\ln(r/0)}{2 \pi K_1 L}$, which is not well-defined for a solid cylinder. Instead, we use the area-weighted average conductivity for parallel heat flow across the cross-section. The effective thermal conductivity $K_{eff}$ is given by $K_{eff} = \dfrac{\sum K_i A_i}{\sum A_i}$.

The area of the inner cylinder is $A_1 = \pi r^2$. The area of the outer shell is $A_2 = \pi (2r)^2 - \pi r^2 = 3 \pi r^2$.

The total area is $A = A_1 + A_2 = 4 \pi r^2$.

Therefore, $K_{eff} = \dfrac{K_1 A_1 + K_2 A_2}{A_1 + A_2} = \dfrac{K_1 (\pi r^2) + K_2 (3 \pi r^2)}{4 \pi r^2} = \dfrac{K_1 + 3 K_2}{4} = \dfrac{1}{4}(K_1 + 3 K_2)$.

Answer: $\dfrac{1}{4}\left(K_1+3 K_2\right)$