b

$\text{Let longest edge is 2L , Shortest edge is L}$.

$\text{Resistivity} = \rho$ ,$Thickness = t$

$\text{Maximum Area} = 2Lt$ , $\text{Minimum area} = Lt$

$ R_{max} = \frac{\rho.L_{max}}{A_{min}} =  \frac{\rho.2L}{L.t} = \frac{2\rho}{t}$

$ R_{max} = \frac{\rho.L_{min}}{A_{max}} =  \frac{\rho.L}{2L.t} = \frac{\rho}{2t}$

$\Rightarrow \frac{R_{max}}{R_{min}} = 4$