Explanation

(b) The temperature of the wire increase to such a value at which the heat produced per second equals heat lost per second due to radiation.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ i^{2}\left( \frac{\rho l}{\pi r^{2}} \right) = H \times 2\pi rl$
Where $H$ is heat lost per second per unit area due to radiation.
Hence, $i^{2} \propto r^{3}$
So, $\frac{i_{1}^{2}}{i_{2}^{2}}\ = \frac{r_{1}^{3}}{r_{2}^{3}}$
Or $r_{2} = r_{1}\left( \frac{i_{2}}{i_{1}} \right)^{\frac{2}{3}}$
$\left\lbrack Here\ :r_{1} = 1\ mm,\ i_{1} = 1.5\ A,\ i_{2} = 3\ A \right\rbrack$
$r_{2} = 1 \times \left( \frac{3}{1.5} \right)^{2/3} = 4^{1/3}\ mm$