Concept
$\Rightarrow$ In this question we have a R/W head on each surface and all the heads Read/Write data from their respective surface simultaneously.
$\textrm{Data tranfer rate}$ = $\frac{\textrm{16 * Total data present on 1 surface of disk}}{\textrm{Time taken to transfer that data}}$
$\textrm{Total data present on 1 surface of disk}$
=$128 \frac{tracks}{surface}256\frac{sector}{track}*512\frac{bytes}{sector}$
= $128* 256 * 512 \frac{bytes}{surface}$
= $2^{24}\frac{bytes}{surface}$
$\textrm{Time taken to transfer that data}$
Disk is rotating at $3600$ revolutions per minute
$\Rightarrow$ $3600$ revolutions in $60$ seconds
$\Rightarrow$ The Disk performs $1$ revolution in $\frac{1}{60}$ seconds
Also, in $1$ revolution a R/W head can read data of $1$ track and a surface has $128$ tracks.
$\Rightarrow$ Time to R/W $1$ surface(time to read $128$ tracks) = $128* \frac{1}{60}$ seconds = $\frac{128}{60}$ seconds
$\Rightarrow$ Time to R/W $16$ surface = $\frac{128}{60}$ seconds ($\because$ $16$ R/W heads are reading data simultaneously from each surface)
$\textrm{Data tranfer rate }$
=$\frac{\textrm{16 * Total data present on a surfaces of disk}}{\textrm{Time taken to transfer that data}}$
= $16 *2^{24} *\frac{60}{128} Bytes/second $
= $ 2^{21}* 60 Bytes/second$
=$2*60 MB/second$
=$120 MB/second$