A -----------4000 KM ------------> B
$T_t = \frac{1000\ bits}{100\ kbps} =10 \ ms$
$T_p = 5 \times 4000 = 20000 \ \mu s = 20 \ ms$
Time to send 1 packet and receive ACK $= 2 \times T_p +T_t = 2 \times 20 +10 = 50\ ms$
as given window size is 3, so we can send 3 packets in 50 ms i.e, 3000 bits
so actual data rate between A & B will be $\frac{3000}{50} = 60\ kbps$
This means B receives data at the rate of 60 kbps, now for no flooding B should also transfer data at the rate of 60 kbps
B -----------1000 KM ------------> C
let, x be the data rate
$T_t = \frac{1000\ bits}{x\ kbps} $
$T_p = 5 \times 1000 = 5000 \ \mu s = 5 \ ms$
Time to send 1 packet and receive ACK $= 2 \times T_p +T_t = 2 \times 5 + \frac{1000}{x} $
Time required to transfer 1 packet $= \frac{1000}{60} = 16.67 \ ms$
$16.67 \ ms = 10+\frac{1000}{x}\\ 6.67=\frac{1000}{x}\\ x=\frac{1000}{6.67}\\ x=150\ kbps$
So, minimum Transmission rate required between B & C is 150 kbps