Given in other part of question, disk makes $5400$ revolution per minute
$60$ sec= $5400$ revolutions
$1$ revolution = $00.11$ sec
Size of track = $50*512$ B = $25600$ B =$25$ KB
If one track of data can be transferred in $1$ revolution that means $25$ KB data is transferred in $0.11$ sec
Data transfer rate(Data transferred/sec) = $25$ KB/$0.11$ = $2272.72$ KB/sec