Speed up is at least 5.
So , $5 \leqslant \frac{(Non pipelined time )}{ Pipeline time}$
Here , target address is available at the 4th clock cycle. So , if we execute any instruction at the 1st clock cycle , we would need to wait till 4th clock cycle to get the output. So, this is stall. So , stall cycle = (4-1) =3
So ,
$5 \leqslant \frac{(6 )}{(1+stall frequency * stall penalty)}$
Now , here question asks about when not branch.
$5 \leqslant \frac{(6 )}{(1+(1-stall frequency) * stall penalty)}$
$5 \leqslant \frac{(6 )}{(1+(1-X) * 3)}$
so , on solving , x comes 0.933
So , 1-x comes as 0.066. So , not to branch probability 0.07.