A frame is $1\text{ KB}$ and takes $\dfrac{8\times 10^3}{1.5\times 10^6}\ s = 5.33\ ms$
to reach the destination. ( $8$ is used to convert byte to bits)
Adding the propagation delay of $50\ ms,$ the total time will be $50 + 5.33 = 55.33\ ms$
Now, we need the $\text{ACK}$ to reach back also, so the time between a packet is sent
and an $\text{ACK}$ is received $= 55.33 + 50\text{(transmission time of ACK neglected)}$
$= 105.33\ ms$
The channel band width is $1.5\text{ Mbps},$ so in $1\ ms,\ 1.5K$ bits can be transferred
and so in $105.33\ ms,\ 157.995 K$ bits can be transferred.
To, ensure $60\%$ utilization, amount of bits to be transferred in
$1\ ms = 157.995\times 0.6 = 94.797\ Kb =\dfrac{ 94.797}{(8\times 1000)}$ frames
$= 11.849\text{ frames}\approx 12\text{ frames}.$
(we bounded up to ensure at least $60\%$ utilization)
So, we need a minimum window size of $12.$
Now, in selective repeat protocol, the window size must be less than half the
sequence number space.
http://stackoverflow.com/questions/3999065/why-is-window-size-less-than-or -equal-to-half-the-sequence-number-in-sr-protocol
So, this means sequence number space must be larger than $2\times 12 = 24.$
To have a sequence number space of $24,$ sequence bits must be at least $\log_{2} 24 = 5$