Explanation

(a) The temperature of the wire increases to such a value at which, the heat produced per second equals heat lost per second due to radiation $ie$,
$I^{2}\left( \ \frac{\rho l}{\pi r^{2}} \right) = H \times 2\pi rl$, where $H$ is heat lost per second per unit area due to radiation.
Hence, $I^{2} \propto r^{3}$
So $\ \ \frac{I_{1}^{2}}{I_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}\ or\ \ r_{2} = r_{1}\left( I_{2}/I_{1} \right)^{2/3}$
$= 1 \times (3.0/1.5)^{2/3} = 4^{1/3}$ mm.