Answer
$$T / m g \approx 0.260$$
Work Step by Step
There are six legs, and the vertical component of the tension force in each leg is
$T \sin \theta$ where $\theta=40^{\circ} .$ For vertical equilibrium (zero acceleration in the $y$ direction) then Newton's second law leads to
$$
6\ T \sin \theta=m g \Rightarrow T=\frac{m g}{6 \sin \theta}
$$
$\text { which (expressed as a multiple of the bug's weight } m g)$ gives roughly $$T / m g \approx 0.260$$
