A $3000\ km$ long trunk operates at $1.536\ mbps$ and is used to transmit $64$ bytes frames and uses sliding window protocol. The propagation speed is $6\mu sec/km$

The transmission and propagation delays are respectively

Given,

Frame size $=64$ bytes

Data rate $=1.536\ mbps$

Distance $=3000\ km$

Propagation delay per km $=6\mu sec/km$

First, convert frame size into bits.

$1$ byte $=8$ bits

So,

$64$ bytes $=64\times 8=512$ bits

Now, transmission delay is given by:

$T_t=\frac{\text{Frame size in bits}}{\text{Data rate}}$

$T_t=\frac{512}{1.536\times 10^6}$ seconds

$T_t=333.33\times 10^{-6}$ seconds

$T_t=333.33\mu sec$

Now, propagation delay is given by:

$T_p=\text{Distance}\times \text{Propagation delay per km}$

$T_p=3000\times 6$

$T_p=18000\mu sec$

Therefore,

$T_t=333.33\mu sec$

$T_p=18000\mu sec$