The fact that you have groups with different sample sizes does not have any direct effect on whether it is reasonable to believe that the populations (e.g., their means) differ. To assess whether you might typically get group sample distributions that differ like yours do if the groups really were identical, we can perform a hypothesis test. A good statistical hypothesis test, such as the $t$-test, will appropriately take the two sample sizes into account. The $t$-test assumes:
- that the data are independent,
(I cannot verify that and the validity of any conclusions based on the given descriptive statistics could be threatened if the data are not independent.)
- that the variances are equal,
(This is not plausible from the information given, but the Welch version of the $t$-test can be applied, so we needn't be too concerned.)
- that the data within each group is normally distributed.
(I can't tell if that's true from what is given, but the point of that assumption is to guarantee that the sampling distributions of the means would be normal, and hence that the sampling distribution of the $t$-statistic will be $t$. Your datasets are so large that that should be the case even if the data are not normal. Thus, we needn't be too concerned here either.)
- The $t$-test does not assume that the sample sizes are the same.
Based on these considerations, we should be comfortable conducting a $t$-test to compare any two of your groups. Here is an example (conducted with R):
library(BSDA) # we need this package for the tsum.test function
# test if C > B
tsum.test(mean.x=6.60, s.x=12.49522047, n.x=320426,
mean.y=3.51, s.y=8.426020509, n.y=10152,
alternative="greater", var.equal=FALSE)
#
# Welch Modified Two-Sample t-Test
#
# data: Summarized x and y
# t = 35.726, df = 11613, p-value < 2.2e-16
# alternative hypothesis: true difference in means is greater than 0
# 95 percent confidence interval:
# 2.947723 NA
# sample estimates:
# mean of x mean of y
# 6.60 3.51