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 $\log_{2} 24 = 5$