A paper dipped in a dil. H_2SO_4 solution of ' X ' upon treatment...

$SO_2$ gas acts as a reducing agent. When a paper dipped in an acidified solution of potassium dichromate ($K_2Cr_2O_7$) is exposed to $SO_2$ gas, the orange-colored dichromate ion ($Cr_2O_7^{2-}$) is reduced to the green-colored chromium(III) ion ($Cr^{3+}$).

The chemical reaction is:
$K_2Cr_2O_7 + H_2SO_4 + 3SO_2 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O$

The formation of chromium(III) sulfate, $Cr_2(SO_4)_3$, is responsible for the green color.

Thus, the compound '$X$' is $K_2Cr_2O_7$.

Answer: $K_2Cr_2O_7$