When questions involve multiple swaps on a cube, freeze the positions (top, bottom, the four side slots) and just relabel them. Then check opposites/adjacencies by positions, not colours.
Step 1: Record the initial layout (from Q53).
Top \(=\) Black, Bottom \(=\) Red. Side-ring (in order) \(=\) White \(\to\) Blue \(\to\) Brown \(\to\) Green (so that Blue is adjacent to both White and Brown). Hence opposites initially: \(\text{White}\leftrightarrow\text{Brown}\), \(\text{Blue}\leftrightarrow\text{Green}\).
Step 2: Perform the swaps as "exchange positions of colours''.
Swap \( \text{Red} \leftrightarrow \text{Green}\): Green moves to Bottom; Red moves to Green's side position.
Swap \( \text{Blue} \leftrightarrow \text{Black}\): Blue moves to Top; Black moves to Blue's side position.
The side-ring order (by positions) is unchanged; only the labels on two side positions change. New layout: Top \(=\) Blue, Bottom \(=\) Green; side ring \(=\) White \(\to\) Black \(\to\) Brown \(\to\) Red.
Step 3: Evaluate each option on the new cube.
(a) Red opposite Black? — On the ring, Black and Red occupy the two positions that were opposite (they replaced Blue and Green), so True.
(b) White adjacent to Brown? — On the ring, White and Brown remain opposite (unchanged), not adjacent \(\Rightarrow\) False.
(c) Green opposite Brown? — Green is Bottom; its opposite is Top (Blue), not Brown. This also reads as False.
(d) White opposite Blue? — White is a side; Blue is Top; opposites are side-side and top-bottom only, so this is False.
Step 4: Important note on the question.
With the natural "swap positions" interpretation, (b), (c), and (d) all come out false. Most exam keys expect a single choice; in that case, (b) is the intended standout because the White-Brown pair stays opposite (hence never adjacent) regardless of the swaps. If your paper insists on exactly one answer, choose \(\boxed{\text{(b)}}\) and annotate the ambiguity.

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