A wire runs from A to B, 0.30 m east, then from B to C, 0.40 m north, then from C to D, 0.30 m west…

Step 1. The whole wire. A is at the start, D is at the end. D is $0.30 - 0.30 = 0$ m east and $0.40$ m north of A. So the effective length is $L_{\text{eff}} = 0.40$ m. $F = BIL_{\text{eff}} = 0.20\ \text{T} \times 4.0\ \text{A} \times 0.40\ \text{m} = 0.32\ \text{N}$. The effective length is north, the field is out of the page, and $\hat{j} \times \hat{k} = \hat{i}$. The force points east.

Step 2. The three parts. AB: $F = 0.20 \times 4.0 \times 0.30 = 0.24$ N, with $\hat{i} \times \hat{k} = -\hat{j}$, to the south. BC: $F = 0.20 \times 4.0 \times 0.40 = 0.32$ N, to the east. CD: $F = 0.24$ N, with $(-\hat{i}) \times \hat{k} = \hat{j}$, to the north. The south and north forces cancel. The east force is $0.32$ N. It agrees.

Step 3. The half circle. $F = BI(2R) = 0.20\ \text{T} \times 5.0\ \text{A} \times 0.30\ \text{m} = 0.30\ \text{N}$.