Concept:
For a cyclic process,
\[
\Delta U=0.
\]
Hence, from the first law of thermodynamics,
\[
Q=W.
\]
Therefore, the heat absorbed by the system is equal to the area enclosed by the cycle on the \(PV\)-diagram.
Step 1: Identify the shape of the cycle.
The cycle is a circle on the \(PV\)-diagram.
From the figure,
\[
P_{\max}=30\,\text{kPa},
\qquad
P_{\min}=10\,\text{kPa}.
\]
Hence the radius along the pressure axis is
\[
r_P=\frac{30-10}{2}=10\,\text{kPa}.
\]
Similarly,
\[
V_{\max}=30\,\text{L},
\qquad
V_{\min}=10\,\text{L}.
\]
Thus,
\[
r_V=\frac{30-10}{2}=10\,\text{L}.
\]
Step 2: Calculate the area enclosed by the cycle.
\[
W=\text{Area}
=
\pi r_Pr_V.
\]
\[
W
=
\pi(10\,\text{kPa})(10\,\text{L}).
\]
\[
W
=
100\pi\,(\text{kPa}\cdot\text{L}).
\]
Step 3: Convert into joules.
Since
\[
1\,\text{kPa}\cdot\text{L}
=
10^3\,\text{Pa}\times10^{-3}\,\text{m}^3
=
1\,\text{J},
\]
we get
\[
W=100\pi\,\text{J}.
\]
\[
W=10^2\pi\,\text{J}.
\]
Step 4: Write the final answer.
\[
\boxed{Q=W=10^2\pi\,\text{J}}
\]
\[
\boxed{\text{Answer = (C)}}
\]