Let P0's BT be x units and P2's BT be y units.
Now, when P0 has finished 6 units, means left is x-6. But at this moment P2 has been given next chance. Since it is SRTF, so P2 getting chance/priority more than P0 means P2's BT must have been less than the remaining BT of P0. Otherwise P0 would have continued.
So, y< x-6 ....(i)
Now, P2 executed for 4 units after which P1 came and it was given the CPU time. P1 takes 5 units.
This means P1's BT< the remaining BT of P2. Otherwise P2 would have continued.
The remaining BT of P2 is y-4.
So, 5< y-4.
y> 9 hence min. Value of y is 10.
From equation (i),
y<x-6 => 10<x-6 => x> 16 => min(x)= 17.