Part 1 — bits per sample. In binary PCM the L quantisation levels are addressed by n bits:

\(L = 2^{n} \Rightarrow n = \log_2 L\)

\(n = \log_2 64 = 6\ \text{bits}\)

So each sample is encoded in 6 bits. (The distractor 32 is L/2 — the number of levels per polarity — not a bit count; a bit count must be a small integer such that 2ⁿ = L.)

Part 2 — output SNR of a µ-law companded system.

\(\left(\dfrac{S}{N}\right)_{dB} = 6n + 4.77 - 20\log_{10}\left[\ln(1+\mu)\right]\)

Step A — the companding penalty term.

\(\ln(1+\mu) = \ln(101) = 4.615\)

\(20\log_{10}(4.615) = 20 \times 0.6644 = 13.28\ \text{dB}\)

Step B — substitute n = 6.

\(\left(\dfrac{S}{N}\right)_{dB} = (6\times 6) + 4.77 - 13.28 = 36 + 4.77 - 13.28\)

\(\left(\dfrac{S}{N}\right)_{dB} = 27.49\ \text{dB}\)

(Using the more precise 6.02 dB per bit instead of 6 gives 27.6 dB — the same answer to the accuracy of the options.)

What the terms mean. The 6n term is the familiar "6 dB per bit" rule: every extra bit halves the step size and so doubles the SNR voltage ratio. The −13.28 dB is the price paid for companding: the µ-law compressor deliberately coarsens the steps for large signals so that small signals get fine steps, which sacrifices peak SNR but makes the SNR nearly independent of signal level — the whole point of companding in telephony. Larger µ means more uniform SNR across levels but a bigger penalty at full scale.

Note on the bandwidth given. B = 4 kHz is not needed for either part; it would fix the sampling rate (\(f_s \ge 2B = 8\ \text{kHz}\)) and hence the bit rate \(R_b = n f_s = 48\ \text{kbps}\) — exactly the reasoning behind the 64 kbps standard PCM channel at 8 bits/sample.

Hence, the encoding uses 6 bits and the output SNR is 27.49 dB.