b
\(\begin{array}{l}
As\,we\,know,coefficient\,of\,friction\,\mu  = \frac{F}{N}\\
 \Rightarrow \mu  = \frac{{ma}}{{mg}} = \frac{a}{g}\left( {a = 7.35m{s^{ - 2}}\,given} \right)\\
\therefore \mu  = \frac{{7.35}}{{9.8}} = 0.75
\end{array}\)