#### Answer

186 Joules

#### Work Step by Step

We know the following equation:
$P_B=P_A(\frac{V_A}{V_B})^{\gamma}$
Plugging in the known values, we obtain:
$P_B=696 \ kPa$
Now that we know this, we can break the work up into two paths. The work done overall is given by:
$W=\frac{P_BV_B-P_AV_A}{\gamma-1}-nrTln(\frac{V_A}{V_C})$
Plugging in the known values gives:
$\fbox{W=186 Joules}$