may be this very easy. because of lack of maths skills I'm posting this.
ex: a has 710444 computers and 23461 cables b has 1047577 computers and 39211 cables...........
a 710444 23461
b 1047577 39211
c 247186 8792
d 518264 13886
e 116696 2045
f 1044267 42573
g 470177 18562
h 499851 16525
i 444441 10608
j 981386 40677
k 732669 27183
l 1143795 37862
m 305696 12324
n 106570 4422
o 381875 12230
p 1478656 31646
q 327575 9624
How can I calculate who holds highest number of cables. generally "f". But as you can see bias in number of computers and thus difference in the cables. Is there any way to calculate remove this bias calculate who has highest cables ?
You can divide the number of cables (second column) by the number of computers (first column). That gives you the ratio $$\frac{\text{cables}}{\text{computers}}.$$
Doing the division tells you the number of cables per computer.
Here are couple of results, so that you will know you are doing things right.
(a) $0.033023$
(b) $0.0374302$
(c) $0.0355684$
(d) $0.0267933$
So far, (b) has the highest ratio of cables to computers. The division compensates for the disparities in the numbers of computers. The numbers we get are kind of ugly looking. So you might want to multiply my numbers by $1000$, round them suitably, and label the new column "Number of cables per $1000$ computers." It looks as if rounding to $1$ decimal place would be plenty good enough.
Another approach is to divide the number of computers by the number of cables. If "big" ratios are good with the first approach, then "small" ratios are good with the second approach. For example, in (a) we would have roughly $30.3$, for $b$ roughly $26.7$, and so on. This approach gives you a sorting opposite to the order given by the first approach.
Choose whichever version you feel is more informative.