A horizontal copper rod of mass 12 g and length 0.2 m lies east to west in a horizontal field of…

Step 1. The weight. $mg = 0.012\ \text{kg} \times 9.8\ \text{m/s}^2 = 0.1176\ \text{N}$.

Step 2. The current. $I = \dfrac{mg}{BL} = \dfrac{0.1176\ \text{N}}{0.15\ \text{T} \times 0.2\ \text{m}} = 3.92\ \text{A}$.

Step 3. The direction. Left hand: forefinger north (field), thumb up (force). The middle finger points east.