A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is -180 J. The gas absorbs 250 J of heat along the path ab and 60J along the path bc. The work done by the gas along the path abc is 

Key idea in p-V curve, work done dW, change in internal energy $\mathrm{\Delta U}$ and heat absorbed $\mathrm{\Delta Q}$ are connected with first law of thermodynamics, i.e. $\mathrm{\Delta Q}=\mathrm{\Delta U}+\mathrm{dW}$
And total change in internal energy in complete cycle is always zero, using this equation in different part of the curve, we can solve the given problem.
In process $\mathrm{a}\to \mathrm{b}$ Given ${\mathrm{\Delta Q}}_{\mathrm{ab}}=250\mathrm{J}={\mathrm{\Delta U}}_{\mathrm{ab}}+{\mathrm{dW}}_{\mathrm{ab}}$

 In process $\mathrm{b}\to \mathrm{c}$ Given ${\mathrm{\Delta Q}}_{\mathrm{bc}}=60\mathrm{J}$

Also, V is constant, so $\mathrm{dV}=0;{\mathrm{dW}}_{\mathrm{bc}}=\mathrm{p}{\left(\mathrm{dV}\right)}_{\mathrm{bc}}=0$
$60\mathrm{J}={\mathrm{\Delta U}}_{\mathrm{bc}}+0; {\mathrm{\Delta U}}_{\mathrm{bc}}=60\mathrm{J}$ In process $\mathrm{c}\to \mathrm{a}$ Given ${\mathrm{\Delta U}}_{\mathrm{ca}}=-180\mathrm{J}$. 

Now for complete cycle, ${\mathrm{\Delta U}}_{\mathrm{abca}}={\mathrm{\Delta U}}_{\mathrm{ab}}+∆{\mathrm{U}}_{\mathrm{bc}}+{\mathrm{\Delta U}}_{\mathrm{ca}}=0$ 

From Eqs.(iii),(iv) and(v) we get

$$\begin{gathered} {\mathrm{\Delta U}}_{\mathrm{ab}}=-{\mathrm{\Delta U}}_{\mathrm{bc}}-{\mathrm{\Delta U}}_{\mathrm{ca}} \\ {\mathrm{\Delta U}}_{\mathrm{ab}}=-60+180=120\mathrm{J} \end{gathered}$$
From Eq.(ii) we get $250\mathrm{J}=120\mathrm{J}+{\mathrm{dW}}_{\mathrm{ab}}\Rightarrow {\mathrm{dW}}_{\mathrm{ab}}=130\mathrm{J}$
From Eqs.(i) and (vii) we get  Work done by the gas along the path abc, ${\mathrm{dW}}_{\mathrm{abC}}={\mathrm{dW}}_{\mathrm{ab}}+{\mathrm{dW}}_{\mathrm{bc}}=130\mathrm{J}+0$

$\Rightarrow {\mathrm{dW}}_{\mathrm{abc}}=130\mathrm{J}$