Answer
(a) Lower than
(b) II
Work Step by Step
(a) We know that $f_{peak}\propto T$ and $f_A\lt f_B$; therefore, $T_A\lt T_B$. Thus, the temperature of black body A is less than the temperature of black body B.
(b) We know that the best explanation is option (II) -- that is, black body B has the higher temperature because an increase in temperature means an increase in frequency, which corresponds to a decrease in wavelength.
