Answer
1.5 mm.
Work Step by Step
Draw the free-body diagram for the needle, which is similar to Figure 10–36a (page 281). Assume that the surface tension force acts vertically. The needle is of length L.
The needle’s weight equals the net surface tension force.
$$mg=2F_T$$
$$\rho_{needle}\pi(0.5d_{needle})^2Lg=2\gamma L$$
$$ d_{needle}=\sqrt{\frac{8\gamma }{\rho_{needle}\pi g}} $$
$$ d_{needle}=\sqrt{\frac{8(0.072\;N/m) }{(7800\;kg/m^3)\pi (9.8\;m/s^2)}}\approx 1.5\;mm $$