The expression for the magnetic field at the centre of a circular
coil of $n$ turns.

$B=\frac{{\mu }_{0}i}{2\mathrm{r}}n$

If length of the wire is doubled,

$2 \left(2\pi rn\right)=2\pi r^{\text{'}}{n}^{\text{'}}$

$\left(2 r n\right)=\frac{r}{3}{n}^{\text{'}}$

$n^{\text{'}}=6\mathrm{n}$

For the same field at the centre,

$\frac{{\mu }_{0}in}{2\mathrm{r}}=\frac{{\mu }_0{i}^{\text{'}}{n}^{\text{'}}}{2{\mathrm{r}}^{\text{'}}}$

$\frac{{\mu }_{0}in}{2\mathrm{r}}=\frac{{\mu }_{0}i \left(6 \mathrm{n}\right)}{2 \left({\displaystyle \frac{r}{3}}\right)}$

$\Rightarrow i^{\text{'}}=\frac{i}{18}$

Since volume will remain same.

$\therefore l\times A=2l\times A^{\text{'}}$

$\Rightarrow A\text{'}=\frac{A}{2}$

The resistance after doubling the length,

$R^{\text{'}}=\frac{\rho l^{\text{'}}}{A\text{'}}$

$\Rightarrow R^{\text{'}}=\frac{\rho \left(2\mathrm{l}\right)}{\left(A/2\right)}$

$\Rightarrow R^{\text{'}}=\frac{4\mathrm{\rho l}}{A}$

$\Rightarrow R^{\text{'}}=4 \mathrm{R}$

The EMF is,

$E^{\text{'}}=i^{\text{'}}{R}^{\text{'}}$

$\Rightarrow E^{\text{'}}=\left(\frac{i}{18}\right) \left(4 \mathrm{R}\right)=\frac{2}{9}E$