We know that, the resistance of wire is
$$
\mathrm{R}=\rho \frac{l}{\mathrm{~A}}
$$
where $\mathrm{A}$ is cross-sectional area of conductor as given that:
$$
\begin{aligned}
l_1 &=l \\
\mathrm{r}_1 &=\frac{1}{2} \mathrm{~mm} \\
&=0.5 \times 10^{-3} \mathrm{~m} \\
l_2 &=l \\
\mathrm{r}_2 &=\frac{2}{2} \mathrm{~mm} \\
&=1 \times 10^{-3} \mathrm{~m}
\end{aligned}
$$
and
The resistance of first conductor
$$
\begin{aligned}
&\mathrm{R}_{\mathrm{A}}=\frac{\rho l_1}{\pi \mathrm{r}_1^2} \\
&\mathrm{R}_{\mathrm{A}}=\frac{\rho \mathrm{l}}{\pi\left(10^{-3} \times 0.5\right)^2}
\end{aligned}
$$
The resistance of second conductor,
$$
\begin{array}{r}
\mathrm{R}_{\mathrm{B}}=\frac{\rho l_2}{\pi\left(\mathrm{r}_2^2-\mathrm{r}_1^2\right)} \\
\mathrm{R}_{\mathrm{B}}=\frac{\rho l}{\pi\left[\left(10^{-3}\right)^2-\left(0.5 \times 10^{-3}\right)^2\right]}
\end{array}
$$
Now, the ratio of two resistances is
$$
\begin{aligned}
&\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\left(10^{-3}\right)^2-\left(0.5 \times 10^{-3}\right)^2}{\left(0.5 \times 10^{-3}\right)^2}=\frac{0.75}{0.25} \\
&\mathrm{R}_{\mathrm{A}}: \mathrm{R}_{\mathrm{B}}=3: 1 \\
&
\end{aligned}
$$