An ideal gas is taken through a cyclic process A arrow B arrow C...

The net work done in the cyclic process is the sum of the work done in each individual process:
$W_{\text{net}} = W_{AB} + W_{BC} + W_{CA}$

1) Process $A \rightarrow B$ is an isobaric expansion at $P = 400 \text{ kPa}$:
$W_{AB} = P \Delta V = 400 \times 10^3 \text{ Pa} \times (5 - 2) \times 10^{-3} \text{ m}^3 = 1200 \text{ J}$

2) Process $B \rightarrow C$ is a linear process from $V = 5 \text{ L}$ to $V = 2 \text{ L}$:
The work done is the area under the $P-V$ graph (a trapezium):
$W_{BC} = \dfrac{1}{2} (P_B + P_C) (V_C - V_B)$
$W_{BC} = \dfrac{1}{2} (400 + P_C) \times 10^3 \times (2 - 5) \times 10^{-3} = -1.5 (400 + P_C) \text{ J}$

3) Process $C \rightarrow A$ is isochoric ($\Delta V = 0$):
$W_{CA} = 0 \text{ J}$

Equating the sum to the given net work of $-450 \text{ J}$:
$1200 - 1.5(400 + P_C) + 0 = -450$
$1200 - 600 - 1.5 P_C = -450$
$600 - 1.5 P_C = -450$
$1.5 P_C = 1050$
$P_C = 700 \text{ kPa}$

Answer: $700 \text{ kPa}$