The eight-cell group with B = 0 uses minterms 0,1,2,8,9,10,11 and don’t-care 3, giving $\bar B$. A second eight-cell group with C = 1 uses 2,6,10,11 and don’t-cares 3,7,14,15, giving C. These groups cover every required 1 and no required 0. Hence $f=\bar B+C$, option B.
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