CBSE Class 12 · 2008 · SemiconductorsVery Short Answer

Inside the dotted box each input is first inverted and the two inverted signals feed a NOR gate, giving

so the box is an AND gate.
Since $Y=A\cdot B$, the output is high only where both $A$ and $B$ are high.
From the waveforms $A$ is high during $0\text{–}1,3\text{–}5,7\text{–}8$, while $Y$ is high only during $3\text{–}5$ and $7\text{–}8$.
Hence $B$ must be high exactly over those output-high intervals.
$\boxed{\text{AND gate; B high during }t=3\text{–}5\text{ and }t=7\text{–}8.}$