$V_{2}, I_{2}$ be the voltage and current for Nichrome wire of length $L$ and cross-sectional area $2 A$

According to Kirchhoffs law, $I_{1}=\frac{\left(V_{1}-V\right)}{R_{1}}$

$I_{2}=\frac{\left(V-V_{2}\right)}{R_{2}}$

where $\mathrm{R}$ is thee resistance of Nicrome wire, $R=\frac{\rho l}{A}$

Hence,

$R_{1}=\frac{\rho 2 L}{A}=2 R$

$R_{2}=\frac{\rho L}{2 A}=0.5 R$

since current entering from one end leaves from other end,

$I_{1}=I_{2}$

$\frac{8-V}{2}=\frac{V-1}{0.5}$

$V=\frac{6}{2.5}=2.4 V$