6. For the following, assume that no data compression is done; this
would in practice almost never be the case. For (a)-(c), calculate the
bandwidth necessary for transmitting in real time.
a) Video at a resolution of 640 x 480, 3 bytes/pixel, 30 frames/second.
b) 160 x 120 video, 1 byte/pixel, 5 frames/second.
c) CD-ROM music, assuming one CD holds 75 minutes? worth and takes 650 MB.
d) Assume a fax transmits an 8 x 10-inch black-and-white image at a
resolution of 72 pixels per inch. How long would this take over a
14.4-Kbps modem?

If this is a homework question, don't read any further.  That being said:

A:
(3 bytes) for each of (640x480 pixels) for each of (30 frames) per second
(3 bytes) for (307200 pixels) for (30 frames) per second
(921600 bytes) for (30 frames) per second
27648000 bytes per second
221184000 bits per second (bytes * 8)

B:
Same process:

1 byte for (160x120 pixels) for (5 frames) per second
96000 bytes per second
768000 bits per second

C:
650 megs per 75 minutes
650 megs per 4500 seconds
0.1444... megs per second
147.9111... kilobytes per second
151460.9777... bytes per second
1211687.8222... bits per second.

D:
8 inches x 10 inches
8*72 pixels x 10*72 pixels
576 pixels x 720 pixels
414720 pixels
...being black and white (not grayscale), that's 1 bit per pixel:
414720 bits

414720 bits over 14400 bits per second modem (bits/bits per second) = 28.8 seconds.

But then, that's how long it takes over a 14.4K *connection*.  With a
14.4 modem, there's likely to be hardware compression, and the fax
protocols may also employ compression, or run-length encoding.  Also,
for all of the answers, you use the subject "computer networks".  A
computer network that claims a specific bandwidth doesn't include the
protocols you put over the network.  Some networks have a large
overhead to just sending packets, and others have large windowing,
which causes slow back-and-forth communication even if
single-direction streams are blazing fast.