Answer
$Q=900 cal$
Work Step by Step
$\Delta E_{int}=nC_V\Delta T$
We plug in the known values to obtain:
$\Delta E_{int}=3.0(6.00)(50)=900 cal$
We also know that the work done is zero for constant volume processes.
Therefore, from the first law of thermodynamics;
$Q=\Delta E_{int}+W=900+0=900 cal$
