Normally; $T_{p}$ is directly what's given in the question, and $RTT = 2T_p$
But in satellite communnications, $T_p$ is actually twice the value of what is mentioned in the question; because the distance they give is from one end to the satellite — we've to take into account the distance from the satellite to the other end, as well.
And understandably, $RTT = 4T_p$ for satellite communications.
We know efficiency (or utilization here) = $\frac{W_{s}}{1+2a}$
$=> \frac{W_{s}}{1+2a}=\frac{1}{4}$
Now, $a = \frac{T_p}{T_t}$ $= \frac{{\color{Red} 2}dB}{vL}$ = $= \frac{ 2*36504*10^3*10^6 }{3*10^8L}$ = $244360/L$
So,
$4W_s = 1+2(244360/L)$
$=> 507 = 488720/L$
$=> L = 488720/507 = 963.944 bits = 120 Bytes$