Answer
$r_1=1.55mm$
Work Step by Step
As $\rho=\frac{E}{J}$
or $E=\rho{J}$
$E=\rho\frac{i}{A}$
$E=\rho\frac{i}{\pi{r^2}}$
We know:
$E_1=\rho\frac{i}{\pi{r_1^2}}$ .....eq(1)
and $E_3=\rho\frac{i}{\pi{r_3^2}}$ ..........eq(2)
dividing eq(1) by eq(3), we get
$\frac{E_1}{E_3}=\frac{r_3^2}{r_1^2}$
$r_1=r_3\sqrt\frac{E_3}{E_1}$
We plug in the known values to obtain:
$r_1=2\sqrt\frac{1.5}{2.5}$
$r_1=1.55mm$
