Answer
The diameter of the wire is approximately $24\mu m$
Work Step by Step
Power loss across the tungsten wire can be taken by the equation,
$$P=VI$$
The voltage drop across the tungsten wire can be taken by the equation,
$$V=IR$$
Taking $I$ as a common value these two equations can be equalized as
$$\frac{P}{V}=\frac{V}{R}$$
Substitute R in above for $\frac{\rho\times l}{A}$
$\frac{\rho\times l}{A}=\frac{V^{2}}{P}$
$π\times\frac{d^{2}}{4}=\frac{P\times\rho\times l}{V^{2}}$
$d=\sqrt \frac{4P\times\rho\times l}{πV^{2}} = \sqrt \frac{400W\times\ 9.0\times 10^{-7}Ωm\times 0.07m}{π\times120^{2}}$
$d = 2.36 \times 10^{-5}m \approx 24\mu m$
