Hypothesis test shows a significant difference in variances. Is the difference is big or small? Two sample test of hypothesis for comparing two variances shows the following result:
Null hypothesis         σ(Param A) / σ(Param B) = 1
Alternative hypothesis  σ(Param A) / σ(Param B) ≠ 1
Significance level      α = 0.05

Statistics
                                          95% CI for
Variable            N  StDev  Variance      StDevs
Param A         47091  0.100     0.010  (0.100, 0.101)
Param B         47091  0.102     0.010  (0.101, 0.103)

Ratio of standard deviations = 0.982
Ratio of variances = 0.964

95% Confidence Intervals
                            CI for
         CI for StDev      Variance
Method       Ratio           Ratio
Bonett  (0.973, 0.991)  (0.947, 0.982)
Levene  (    *,     *)  (    *,     *)

Tests
                         Test
Method  DF1    DF2  Statistic  P-Value
Bonett    1      —      15.89    0.000
Levene    1  94180       3.15    0.076

Based on the -
p-Value, I can conclude that there is a significant difference.
How do I conclude if the difference is significantly big or small?  
I guess I am missing a theory here, 
I would appreciate if someone leads me to the right way of solving this problem.
 A: Assuming that the Bonett test is applicable, you are correct that the small P-value 0.000 (meaning $<0.0005$) indicates that the two population SDs differ "significaantly" (their ratio differs from 1).
The corresponding 95% CI says that $0.973 \le \frac{\sigma_1}{\sigma_2} \le 0.991,$ which indicates a small difference between the two population SDs. Notice that the CI does not contain $1.$
More directly, the sample SDs are $S_1 = 0.100$ and $S_2 = 0.102.$
The population variances in your computer output are rounded too much so that
there does not appear to be any difference between them. If the sample SDs
are correct to three places, then the sample variances are a little different: $S_1^2 = 0.0100$ and $S_2^2 = 0.0104.$
The very large sample sizes have enabled you to detect, as statistically
significant, a very small difference between the two sample SDs.
Whether the small difference between $S_1 = 0.100$ and $S_2 = 0.102$ is of any
practical importance is not really a statistical question. That is for
the experimenters to decide, based on their knowledge of the data and
what is being measured.
Addendum per Comments:
If you are testing $H_0: \sigma_1^2/\sigma_2^2 = 1$ against 
$H_a: \sigma_1^2/\sigma_2^2 \ne 1$ at the 5% level, using $F = S_1^2/S_2^2$
as the test statistic, then you will reject if $F < 0.9868$ or $F > 1.0182$.
For your data $F = 0.9612 < 0.9868,$ so you reject because $F$ is below
the lower critical value. Below is output from R statistical software for $F$ based on the computer
printout above and a computation of the critical values. I show output from
software because F values for such large samples as yours are not explicitly
printed in tables of the F distribution.
f = (.100/.102)^2;  f
## 0.9611688
qf(.025, 47090, 47090)   # lower critical value
## 0.982098
qf(.975, 47090, 47090)   # upper critical value
## 1.018228

