Let Speed of train= v km/hour

Time taken to complete journey= t hour

$Time=\frac{Distance}{Speed}$

$v \times t =Distance$

$v \times t =528 \dots\dots(1)$

Given that if speed increased by $\frac{11}{2} km /hour$ from its usual speed then train takes $\frac{4}{3}$ hour less.

$(v+\frac{11}{2}) \times (t-\frac{4}{3}) =528 \dots\dots(2)$

$(vt-\frac{4}{3}v+\frac{11}{2}t-\frac{44}{6}) =528 $

From (1)

$(528-\frac{4}{3}v+\frac{11}{2}t-\frac{44}{6}) =528 $

$(-\frac{4}{3}v+\frac{11}{2}t-\frac{44}{6}) =0$

$33t-8v=44$

$33\times \frac{528}{v}-8v=44$

$8v^{2}+44v-17424$

v=44 , $-\frac{99}{2}$

Hence,Option (a)44 is the correct choice.