The partitions are $4k, 8k, 20k, 2k$, now due to the best-fit algorithm,
- Size of $2k$ job will fit in $2k$ partition and execute for $4$ unit
- Size of $14k$ job will be fit in $20k$ partition and execute for $10$ unit
- Size of $3k$ job will be fit in $4k$ partition and execute for $2$ unit
- Size of $6k$ job will be fit in $8k$ partition now execute for $1$ unit. All partitions are full.
And next job size of $10 k \;(5)$ wait for the partition of $20k$ and after completion of no. $2$ job, job no. $5$ will be executed for $1$ unit $(10\;\text{to}\; 11).$ Now, $20 k$ is also waiting for a partition of $20k$ because it is the best fit for it. So after completion of job $5$, it will be fit. So, it will execute for $8$ unit which is $11$ to $19$. So, at $19$ unit $20k$ job will be completed.
The answer should be $19$ units.