The work done by gas in process $ABCDEFGHA$ is $X{{P}_{0}}{{V}_{0}}$, then $X$ is:

Hint: To solve this problem, we can divide the process into parts and by identifying the processes as isochoric or isobaric we can find out the work done by the gas in each of the parts as mentioned in the question. Then finally we can sum up all the work done by each part to get our final answer.

Complete answer:
Let us first break down the process into small parts $A-B-C-D-E-F-G-H-A$ and then consider the first $AB$. As we can see from the diagram that the pressure remains constant throughout the process, hence it is an isobaric process and the work done by the gas in this part will be:
$\begin{align}
  & {{W}_{AB}}={{P}_{0}}\left( {{V}_{0}}-3{{V}_{0}} \right) \\
 & \Rightarrow {{W}_{AB}}=-2{{P}_{0}}{{V}_{0}} \\
\end{align}$
If we consider part $BC$, there is no change in volume, hence the process is isochoric and the work done by the gas will be:
${{W}_{BC}}=0$
The pressure remains constant throughout the process $CD$, hence the work done by the gas in this part will be:
\[\begin{align}
  & {{W}_{CD}}=4{{P}_{0}}\left( 4{{V}_{0}}-{{V}_{0}} \right) \\
 & \Rightarrow {{W}_{CD}}=4{{P}_{0}}\left( 3{{V}_{0}} \right) \\
 & \Rightarrow {{W}_{CD}}=12{{P}_{0}}{{V}_{0}} \\
\end{align}\]
If we consider part $DE$, there is no change in volume, hence the work done by the gas will be:
${{W}_{DE}}=0$
Similarly, the pressure remains constant throughout the process $EF$, hence the work done by the gas in this part will be:
$\begin{align}
  & {{W}_{EF}}=2{{P}_{0}}\left( 2{{V}_{0}}-4{{V}_{0}} \right) \\
 & \Rightarrow {{W}_{EF}}=4{{P}_{0}}{{V}_{0}} \\
\end{align}$
If we consider part $FG$, there is no change in volume, hence the work done by the gas will be:
${{W}_{FG}}=0$
The pressure remains constant throughout the process $GH$, hence the work done by the gas in this part will be:
$\begin{align}
  & {{W}_{GH}}=3{{P}_{0}}\left( 3{{V}_{0}}-2{{V}_{0}} \right) \\
 & \Rightarrow {{W}_{GH}}=3{{P}_{0}}{{V}_{0}} \\
\end{align}$
The work done by the gas in process $HA$ will be:
${{W}_{HA}}=0$
The total work done by the gas will be:
$\begin{align}
  & {{W}_{ABCDEFGHA}}=-2{{P}_{0}}{{V}_{0}}+0+12{{P}_{0}}{{V}_{0}}+0-4{{P}_{0}}{{V}_{0}}+0+3{{P}_{0}}{{V}_{0}} \\
 & {{W}_{ABCDEFGHA}}=9{{P}_{0}}{{V}_{0}} \\
\end{align}$
Hence, the value of $X$ is $9$.

Note:
The work done in an isochoric process is zero because the work done by the gas is the product of pressure and the change in volume of the gas. Since, volume remains constant in an isochoric process, hence the change in volume is zero and eventually the work done by the gas is also zero.