two metallic conductors Aand B are …

Resistance of wire {tex}A({R_1}) = \frac{{\rho l}}{A} = \frac{{\rho l}}{{\pi {r^2}}}{/tex}
Resistance of wire {tex}B({R_2}) = \frac{{\rho {l^/}}}{{{A^/}}} = \frac{{\rho 2l}}{{\pi (2{r})^2}} = \frac{{\rho 2l}}{{4\pi {r^2}}}{/tex}
Total resistance in series
R={tex}{R_1} + {R_2}{/tex}
R = {tex}\frac{{\rho l}}{{\pi {r^2}}} + \frac{{\rho 2l}}{{4\pi {r^2}}}{/tex}
R = {tex}\frac{{\rho l}}{{\pi {r^2}}}(1 + \frac{1}{2}) = \frac{{3\rho l}}{{2\pi {r^2}}}{/tex}
Ratio of the total resistance in series to the resistance of A =
{tex}\frac{R}{{{R_1}}} = \frac{{3\rho l}}{{2\pi {r^2}}}/\frac{{\rho l}}{{\pi {r^2}}}{/tex}
{tex}\frac{R}{{{R_1}}} = \frac{{3\rho l}}{{2\pi {r^2}}} \times \frac{{\pi {r^2}}}{{\rho l}}{/tex}
{tex}\frac{R}{{{R_1}}} = \frac{3}{2}{/tex}

So, the required answer =3:2