If resistivity of all four wire connected in parallel is different as given in figure and all other dimensions of wire are same then current $i$ is (resistance of wire having resistivity $\rho $ is $R_0$ )
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$\mathrm{R}_{0}=\frac{\rho \mathrm{L}}{\mathrm{A}}=\frac{1}{\mathrm{R}}=\frac{1}{\mathrm{R}_{0}}+\frac{1}{2 \mathrm{R}_{0}}+\frac{1}{4 \mathrm{R}_{0}}+\frac{1}{6 \mathrm{R}_{0}}$
$=\frac{12+6+3+2}{12 R_{0}}$
$R=\frac{12 R_{0}}{23} \quad i=\frac{\varepsilon}{\left(\frac{2 R_{0}}{23}\right)}=\frac{23 \varepsilon}{12 R_{0}}$
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