As material of rod is not changed, resistivity of both rods is same.

Also, volume of material is same for both rods, so

$A_1 l_1=A_2 l_2$ $\text { or } A_1 L=A_2(2 L)$ $\Rightarrow A_2=\frac{A_1}{2}$

Now, using $R=\rho \frac{l}{A}$, we have

$R_2=\rho \frac{2 L}{\left(A_1 / 2\right)}=4\left(\frac{\rho L}{A_1}\right)$

or $R_2=4 R$