Answer
$\Delta E_{int} = 0 J $
Work Step by Step
Since the gas is an ideal gas, the work done for isothermal process in path $1 \rightarrow 2$ is just the same as the heat, Q during that path according to the 'first law of thermodynamics'.
$\Delta E_{int} = W - Q$ Take w and Q from (a) and (d),
$\Delta E_{int} = 700 J - 700 J$
$\Delta E_{int} = 0 J $
