Discussion
The same frame's individual bit duration is
Answer: B
Dividing the slot among its eight bits:
\[ \frac{3.9}{8} = 0.49 \ \mathrm{\mu s} \]
giving 0.49 microseconds, which corresponds to the 2.048 megabit aggregate rate as it must. The other options are the slot, the frame and an unrelated round number. Checking that a computed bit interval agrees with the aggregate rate obtained independently is a cheap and effective way to catch an arithmetic slip in a multiplex calculation. The parameter that varies and the parameter that does not are worth naming before any substitution is attempted.
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