A rod PQ of length 20.0 cm and mass 200 g is suspended horizontally...

Given mass of the rod $m = 200 \text{ g} = 0.2 \text{ kg}$ and length $L = 20.0 \text{ cm} = 0.2 \text{ m}$.

The weight of the rod acts downwards:
$W = mg = 0.2 \times 9.8 = 1.96 \text{ N}$

The current $I = 2.0 \text{ A}$ flows from $P$ to $Q$ (rightwards) and the magnetic field $B = 0.500 \text{ T}$ points out of the page. Using the right-hand rule, the magnetic force on the rod is directed downwards:
$F_m = ILB = 2.0 \times 0.2 \times 0.500 = 0.2 \text{ N}$

Total downward force on the rod is:
$F_{\text{net}} = W + F_m = 1.96 + 0.2 = 2.16 \text{ N}$

Since $OP = OQ = PQ = 20.0 \text{ cm}$, $\triangle OPQ$ is an equilateral triangle. The angle each thread makes with the vertical is $30^\circ$.

Let $T$ be the tension in each thread. Balancing the vertical forces:
$2T \cos(30^\circ) = F_{\text{net}}$

$2T \left(\dfrac{\sqrt{3}}{2}\right) = 2.16$

$T = \dfrac{2.16}{\sqrt{3}} \approx 1.247 \text{ N}$

Rounding off, we get $T \approx 1.25 \text{ N}$.