Answer
$\frac{\Delta E_{int} }{p_oV_o} = 6.0 $
Work Step by Step
Process $b \rightarrow c$ does not involve any volume change but the pressure increases, This is an isochoric process. To find the $\frac{\Delta E_{int}}{p_oV_o} $, we must know the $Q$ in terms of $p_oV_o$.
$Q = \frac{3}{2} nR (T_c - T_b)$
Now we need to find $T_c$ and $T_b$ in terms of $p_oV_o$.
Note that $p_c = 2.0p_o$ and $V_1 = 4.0 V_o$
$T_c = \frac{p_CV_c}{nR} $
$T_c = \frac{(2.0p_o)(4.0 V_o)}{nR} $
$T_c = \frac{8.0p_oV_o }{nR}$
and
$T_b = \frac{p_CV_c}{nR} $
$T_b = \frac{(p_o)(4.0 V_o)}{nR} $
$T_b = \frac{4.0p_oV_o }{nR}$
Now substitute into heat energy equation.
$Q = \frac{3}{2} nR (\frac{8.0p_oV_o }{nR} - \frac{4.0p_oV_o }{nR})$
$Q = \frac{3}{2} nR (\frac{4.0p_oV_o }{nR})$
$Q = 6.0p_oV_o $
The first law of thermodynamics goes $\Delta E_{int} = Q - W$ and W = 0 in isochoric. So, $\Delta E_{int} = Q$
$\Delta E_{int} =6.0p_oV_o $
$\frac{\Delta E_{int} }{p_oV_o} = 6.0 $
