Three metallic bars A, B and C are arranged as shown in the figure. Number density of free electrons in the three bars are in ratio, $N_A:N_B:N_C=1:2:3,$  ratio of resistivities of the bar is  ${\rho }_A:{\rho }_B:{\rho }_C=2:3:1$ , lengths are in ratio  $I_A:I_B:I_C=2:2:3$  and radii are in the ratio $r_A:r_B:r_C=1:2:3$  as shown.

 ${\text{P = Power = i}}^{\text{2}}{\text{R = i}}^{\text{2}}\cdot \frac{\text{ρI}}{{\text{πr}}^{\text{2}}}$
As, i through all the wires is same.
$$\begin{gathered} \Rightarrow \text{ P}\propto \frac{\text{ρI}}{{\text{r}}^{\text{2}}} \\ \Rightarrow P_A:P_B:P_C=\frac{{\rho }_{A^{\text{'}}A}}{r_A^2}:\frac{{\rho }_{B^{\text{'}}B}}{r_B^2}:\frac{{\rho }_{C^{\text{'}}C}}{r_C^2}=4:\frac{3}{2}:\frac{1}{3}=24:9:2 \\ \Rightarrow {\text{ P}}_A=24P_0,P_B=9P_0,P_C=2P_0 \\ \therefore \sqrt{P_B{P}_C}={\text{Geometric mean of P}}_{B}and{\text{ P}}_C \\ =\sqrt{{\text{9P}}_{\text{0}}{\text{2P}}_{\text{0}}}=\sqrt{18}{\text{P}}_{\text{0}} \\ \Rightarrow \frac{{\text{P}}_{\text{A}}}{\sqrt{{\text{P}}_{\text{B}}{\text{P}}_{\text{C}}}}=\frac{24{\text{P}}_{\text{0}}}{\sqrt{{\text{18P}}_{\text{0}}}}=4\sqrt{2} \end{gathered}$$
By ohm’s law in microscopic form.
$$\begin{gathered} J=\sigma E \\ \Rightarrow \frac{i}{\pi r^2}=\frac{E}{\rho }\Rightarrow E\propto \frac{\rho }{r^2} \\ \Rightarrow E_A:E_B:E_C=\frac{{\rho }_A}{r_A^2}:\frac{{\rho }_P}{r_B^2}:\frac{{\rho }_C}{r_C^2}=\frac{2}{4}:\frac{3}{4}:\frac{1}{9}=72:27:4 \\ \Rightarrow E_A=72E_0,E_B=27E_0,E_C=4E_0 \\ \Rightarrow \frac{E_B^2}{E_A{E}_C}=\frac{{27}^2}{72\times 4}=\frac{81}{32} \\ \Rightarrow {\text{ 32E}}_{\text{B}}^{\text{2}}=81{\text{E}}_{\text{A}}{\text{E}}_{\text{C}} \end{gathered}$$ 
Drift speed v is related to current i by the relation,
 $\text{v}=\frac{i}{\text{NeA}}=\frac{i}{\text{Ne}\pi r^2}\Rightarrow v\propto \frac{1}{{\text{Nr}}^2}$
 ${\text{V}}_{\text{A}}{\text{:V}}_{\text{B}}{\text{:V}}_{\text{C}}=1:\frac{1}{8}:\frac{1}{27}=216:27:8$
Also, potential difference,
 $\text{V}=IR=\frac{I\text{ρ}I}{\pi r^2}\Rightarrow V\propto \frac{\text{ρ}I}{r^2}$
 $\therefore {\text{V}}_{\text{A}}{\text{:V}}_{\text{B}}{\text{:V}}_{\text{C}}=4:\frac{3}{2}:\frac{1}{3}=24:9:2$