Speedup = $\frac{Time_{without pipeline}}{Time_{pipeline}}$
For 200 instructions, Timewithout pipeline = Number of instructions * Number of stages * TCLK
= 200 * 6 *TCLK
Timewith pipeline = [(Number of non branch instructions * CPInon branch) + (Number of branch instrutions * CPIbranch )] * TCLK
= [x * 1 + (1 - x) * 4] * 200 *TCLK
$\because$ Next instruction address is know after 4th stage. So, stall cycles for each branch instruction is 3 and hence CPI for branch instructions is 4. CPI for non branch instruction is 1.
So, $\frac{200 * 6}{[x * 1 + (1 - x) * 4]*200}$ $\geqslant 5$
Solving, x = 0.93