In Programmed I/O, the CPU issues a command and waits for I/O operations to complete.
So here, CPU will wait for $1\text{ sec}$ to transfer $10\ KB$ of data.
The minimum performance gain for interrupt mode happens for the smallest unit of data transfer – which here is $1$ byte.
Time to transfer $1$ byte of data in programmed I/O mode $=\dfrac{1}{10\; KBps} = 100 \mu s$
In Interrupt mode, to transfer $1$ byte of data, overhead is $4 \times 10^{-6}s = 4\mu s$
Performance gain $=\dfrac{100}{4}= 25$
Thus, (b) is correct answer.