Answer
(a) $4.3 \times 10^6 ~bits/s$
(b) 67% of the bits on the CD are used for encoding and error-correction.
Work Step by Step
(a) $N = \frac{1.2 ~m/s}{0.28\times 10^{-6} ~m/bit} = 4.3 \times 10^6 ~bits/s$
The CD players reads $4.3 \times 10^6 ~bits$ every second.
(b) We can calculate the percentage of bits which are used for encoding and error-correction.
$\frac {N-N_0}{N} \times 100\% =
\frac{4.3 \times 10^6 ~bits/s - 1.4 \times 10^6 ~bits/s}{4.3 \times 10^6 ~bits/s} \times 100\% = 67\%$
67% of the bits on the CD are used for encoding and error-correction.
