for maximum utilization $\eta = 1$,
$\begin{align*} \eta &= SWS \times \frac{\text{Transmission Time}}{\text{Transmission Time + 2$\times$Propagation Time}} \\ \\ 1 &= SWS \times \frac{\text{Transmission Time}}{\text{Transmission Time + 2$\times$Propagation Time}} \\ \end{align*}$
so, $\dfrac{\text{Transmission Time + 2$\times$Propagation Time}}{\text{Transmission Time}} =\text{SWS}$
gives the number of packets that are sent. It means that it is the Sender's Window Size.
But we also know that,
$\text{available sequence numbers} \ \geq \text{ SWS + RWS}$
so, to get the minimum number of bits needed to represent the sequence numbers - we should consider what protocols are in use.
if GBN ARQ:
$\begin{align*} \substack{\large\text{# bits required to rep.} \\ \large\text{ sequence numbers}}
&= \lceil \log_2{(SWS + RWS)} \rceil \\ &= \left \lceil \log_2{\left(\frac{\frac{K}{R} + 2Lt}
{\dfrac{K}{R}}+1 \right)} \right \rceil \\ &= \left \lceil \log_2{\left(\frac{2K + 2LtR}{K}\right)}
\right \rceil \end{align*}$
if Selective Repeat ARQ :
$\begin{align*} \substack{\large\text{# bits required to rep.} \\ \large\text{ sequence numbers}}
&= \lceil \log_2{(2\times SWS)} \rceil \\ &= \left \lceil \log_2{\left(2\times \frac{\frac{K}{R} + 2Lt}{\frac{K}{R}}\right)}
\right \rceil \\ &= \left \lceil \log_2{\left(\frac{2K + 4LtR}{K}\right)} \right \rceil \end{align*}$