Tt= (1500 bytes * 8) / $10^{9}$ = 0.00001.2 seconds = 0.012 milliseconds
utilization = 0.98 = $\frac{W_{s}}{1+2a}$
$a = \frac{t_{p}}{t_{t}}$
(0.012x) / 30.012 = 0.98
0.012x = 29.41176
x = 2450.98
The window size would have to be approximately 2451 packets