When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling, the gas (A) changes into a colourless solid (B). Identify (A) and (B).

In a reaction between acid and salt, a more reactive metal displaces the less reactive metal from the salt and thus it is a type of displacement reaction.
Now, when sulphuric acid was added to salt, it liberated brown gas. The brown-coloured gas represents the evolution of nitrogen dioxide.

${\mathrm{MNO}}_3+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{MHSO}}_4+{\mathrm{HNO}}_3$

$4{\mathrm{HNO}}_3\to 4{\mathrm{NO}}_2+{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}$
 

Thus, (A) is ${\text{NO}}_2$.
The nitrogen gas released intensifies when copper is added:

$3\mathrm{Cu}+8{\mathrm{HNO}}_3\to 3\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+2\mathrm{NO}+4{\mathrm{H}}_2\mathrm{O}$

$2\mathrm{NO}+{\mathrm{O}}_2\to 2{\mathrm{NO}}_2$

${\mathrm{NO}}_2$ gas produced in this reaction is further cooled down to a solid form and while it's conversion the ${\mathrm{NO}}_2$ gas dimerizes to the ${\mathrm{N}}_2{\mathrm{O}}_4$ i.e. nitrogen tetraoxide.
Thus, the colourless solid produced is ${\mathrm{N}}_2{\mathrm{O}}_4$.

Thus, correct answer is ${\mathrm{N}}_2{\mathrm{O}}_4$.