The figure displays the variation of the internal energy U with the...

For an ideal gas, the internal energy $U$ depends only on its temperature $T$.

From the given $U-V$ graph, we can identify the thermodynamic processes:
Process $a \rightarrow b$: $U$ is constant, which implies $T$ is constant. This is an isothermal expansion at $T_a = T_b = 500 \text{ K}$.
Process $b \rightarrow c$: $V$ is constant. This is an isochoric process.
Process $c \rightarrow d$: $U$ is constant, which implies $T$ is constant. This is an isothermal compression at $T_c = T_d = 300 \text{ K}$.
Process $d \rightarrow a$: $V$ is constant. This is an isochoric process.

For a cyclic process, the net change in internal energy is zero ($\Delta U = 0$).
According to the first law of thermodynamics, the net heat absorbed is equal to the net work done:
$Q_{\text{net}} = W_{\text{net}} = W_{ab} + W_{bc} + W_{cd} + W_{da}$

Calculating the work done in each process:
$W_{ab} = nRT_b \ln\left(\dfrac{V_b}{V_a}\right) = nR(500) \ln\left(\dfrac{2V_0}{V_0}\right) = 500 nR \ln 2$
$W_{bc} = 0$ (since volume is constant)
$W_{cd} = nRT_c \ln\left(\dfrac{V_d}{V_c}\right) = nR(300) \ln\left(\dfrac{V_0}{2V_0}\right) = -300 nR \ln 2$
$W_{da} = 0$ (since volume is constant)

Net work done:
$W_{\text{net}} = 500 nR \ln 2 - 300 nR \ln 2 = 200 nR \ln 2$

Substituting the given values ($n = 2.0 \text{ mol}$, $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$, $\ln 2 = 0.693$):
$Q_{\text{net}} = W_{\text{net}} = 200 \times 2.0 \times 8.3 \times 0.693$
$Q_{\text{net}} = 400 \times 8.3 \times 0.693$
$Q_{\text{net}} = 3320 \times 0.693 = 2300.76 \text{ J}$

Rounding to the nearest given option, the net heat absorbed is $2300 \text{ J}$.

Answer: $2300 \text{ J}$