Q: Interpreting a very small odds ratio when the Fisher's exact test is significant I'm having a hard time interpreting odds ratios (OR) for a 2x2 contingency table. 
            No access    Access
Hospital A      0          56
Hospital B      3          18

I already know by a Fisher's exact test that the result is significant (the Fisher exact test statistic value is 0.0182 and it is significant at p < .05).
My understanding is that the chance of having access to the database is of 1 in hospital A and of 18/21=0.8571 in hospital B. The OR would then be of 1.14.
My interpretation would be that there is 1.14 times more chance to have access to the database when you work in the hospital A. 
But 1.14 is a very small difference... can it really be significant? Am I interpreting something the wrong way?

A: First of all, you calculate the Odd Ratio wrong. 
In your case (1)
, OR = ( 56/0 ) / ( 18/3 ) = ( 100% / 0%) / ( 85.71% / 14.29%) = inf

Therefore second, you ask the wrong question. 
The fact is that the OR is so good that even the significant p-value( = 0.0182) is not as good now. 
Hence, as what @whuber commented, the good effect size is only halfway to the test significance. 

A: (Answer from previous comment).
This is an older question, but it appears that the odds ratio (OR) is not calculated correctly in the question. This is mentioned in the answer by @Jason .
Since there is a 0 in the table, likely one could add 0.5 to all cells (Haldane-Anscombe correction), and calculate the OR in the usual manner. This would yield an OR (not log) of c. 21. With this correction, Hospital A has a ratio of c. 100 and Hospital B has a ratio of c. 5.
In R,
library(vcd)

oddsratio(matrix( c(0, 3, 56, 18), nrow=2), log=FALSE)


   ### 0.04677623

1 / 0.04678

   ### 21.37666


A: 
But 1.14 is a very small difference... can it really be significant? Am I interpreting something the wrong way?

In addition to previous answers, perhaps, yes, you are over-interpreting the p-value.
P = 0.0182 is not that small. After all, under the hypothesis of no difference you can get such or a more extreme result 1 in ~55 attempts (1 / 0.0182 = 54.9). Or it is equivalent to seeing 6 heads in a row from a fair coin (-log2(0.0182) = 5.78).
