Finding structure of a compound given its reactions
I have taken the liberty of identifying your unknown compound.  If you'd like to work through the problem yourself, then carefully scroll down to the part of the answer that deals with the iodoform test.  Good luck!

What we know


Just looking at the molecular formula we see that there are too many hydrogens for the 6 carbons to be in a benzene ring.
does not change the colour of acidic dichromate solution - acidic dichromate readily oxidizes primary and secondary alcohols to aldehydes and ketones respectively; tertiary alcohols are not oxidized. The aldehyde produced from a primary alcohol can be further oxidized to a carboxylic acid, this results in a solution color change from orange to green. But there was no color change, therefore there is no primary $\ce{-OH}$ or aldehyde group in compound A.
Compound A on treatment with H2SO4 produces alkene - this is a common reaction that causes deydration (elimination of water) and converts an alcohol into an olefin.  Therefore compound A is (as you suggested) an alcohol (but not a primary alcohol).
oxidative ozonolysis gives a molecule $\ce{C6H10O3}$ - ozonolysis will cleave the double bond in the compound produced upon dehydration.  The key fact here is that the resulting product still contains 6 carbon atoms.  If the double bond involved carbons outside of a ring we would wind up with an ozonolysis product containing fewer than 6 carbons.




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Only if the double bond is contained within a ring will we retain all 6 carbon atoms.  Therefore our double bond must be in a ring.




positive iodoform test - This is our final clue, the iodoform test (or more generally the haloform reaction) is a test that detects the presence of methyl ketones and generates iodoform ($\ce{HCI3}$) in the process.




Therefore we know that there is a methyl ketone in our ozonolysis product.

Combining all of this information (non-aromatic, ring, non-primary alcohol, methyl group) leads to the following structure

(mouse over to see the answer)


   


which is called


   1-methylcyclopentanol.


Note: some methylated-cyclobutanols might also be considered, but when these dehydrate and produce an alkene, the double bond is likely to be placed external to the ring (or [advanced topic] external to a cyclopropane ring formed upon rearrangement) and we would not wind up with 6 carbons in our ozonolysis product.