Gas 'A' undergoes change from state 'X' to state 'Y'. In this...

Using the first law of thermodynamics, $\Delta U = q + w$.

For the process from state X to state Y:
Heat absorbed by the gas, $q_{X \to Y} = +10$ J
Work done by the gas, $w_{X \to Y} = -18$ J
Change in internal energy, $\Delta U_{X \to Y} = q_{X \to Y} + w_{X \to Y} = 10 - 18 = -8$ J

For the reverse process from state Y to state X:
Change in internal energy, $\Delta U_{Y \to X} = -\Delta U_{X \to Y} = +8$ J
Heat evolved by the gas, $q_{Y \to X} = -6$ J
Let the work done on the gas be $w_{Y \to X}$.
$\Delta U_{Y \to X} = q_{Y \to X} + w_{Y \to X}$
$8 = -6 + w_{Y \to X}$
$w_{Y \to X} = +14$ J

Since the work done is positive, $14$ J of work is done on the gas by the surrounding.

Answer: $14$ J of the work is done on the gas 'A' by the surrounding.