Given,

Bandwidth (Bw) $= 100$ Mbps

distance (d) $= 2100$ Km $= 2100 \times 10^3$ m

velocity (v) $= 3 \times 10^8$ m/s

packet size (L) = $1000$ bytes $= 8000$ bits

Here, sender is station and receiver is satellite

So, Total time $=$ Transmission time $+$ Propagation time

Transmission time $= \frac{L}{Bw} = \frac {8000\ b}{100 \times 10^6\ b/sec} = 0.08$ ms

Propagation time $= \frac{d}{v} = \frac {2100 \times 10^3\ m}{3 \times 10^8 \ m/sec} = 7$ ms

$\therefore$ Total time $= 0.08\ + \ 7 = 7.08$ ms