Answer
$0.130mm$
Work Step by Step
We can find the required diameter as follows:
$d=\sqrt{\frac{4\rho_{tungsten}L}{\pi R}}$
We plug in the known values to obtain:
$d=\sqrt{\frac{4\times 5.6\times 10^{-8}\Omega.m\times 1.2m}{\pi \times 5.0\Omega}}$
$d=0.130\times 10^{-3}m$
$d=0.130mm$
