A sample of an ideal gas is taken through the cyclic process ABCA...

In the part $A \to B$, the volume remains constant. Thus, the work done by the gas is zero. It is given that the heat absorbed by the gas is $40J$. From first law of thermodynamics, $Q = \DeltaU + W$

$⇒ 40 = \DeltaU + 0 ⇒ \DeltaU = 40J$

or $\DeltaU = {U}_{B} - {U}_{A} = 40J$

The increase in internal energy from $A$ to $B$ is $40J$. The internal energy is $1560J$ at $A$, then

${U}_{B} - 1560 = 40J ⇒ {U}_{B} = 1600J$

In the part $B \to C$, the work done by the gas is $W = 50J$ and no heat is given to the system. Again using $Q = \DeltaU + W$

$0 = \DeltaU - 50$ or $\DeltaU = {U}_{C} - {U}_{B} = 50J$

${U}_{C} - 1600 = 50J ⇒ {U}_{C} = 1650J$

The change in internal energy from $C \to A$,

$∆ U = 1560 - 1650 = - 90J$

The heat given to the system is $Q = - 60J$

Using $Q = \DeltaU + W$, we get $W = Q - ∆ U = - 60 - \left(- 90\right) = 30J$