BW = 20 Mbps or 20 * $10^{6}$ bits per second
distance = 2400 km or 2400 * $10^{3}$ m
packet size = 2 KB or 2 * $10^{3}$ * 8 bits
speed of light = 3 * $10^{8}$ m/sec
Window size to achieve 100% Link utilization can be calculated through a simple formula:
$w \geq$ 1 + 2a
where a = propagation delay / transmission time
propagation delay = distance / speed of light = (2400 * $10^{3}$) / (3 * $10^{8}$) = 8 ms
Transmission time = packet size / BW = (2 * $10^{3}$ * 8) / (20 * $10^{6}$) = 0.8 ms
$w \geq$ 1 + 2 * (8/0.8) = 1 + 20 = 21.
Sequence numbers = 2 * W = 2 * 21 = 42
To represent 42 sequence numbers in binary, $log_{2}(42)$, 6 bits are needed.
So a minimum of 6 bits is needed for 100% utilization.