$11?$

Edit :

Reasoning :

Every flop will reduce frequency by $2$ and cascaded flipflops are likely to depend on previous one .

$\therefore$ let initial frq $=F$

Then : $\Large\frac{\frac{F}{2^N}}{\frac{F}{2^{\frac{N+1}{2}}}}$ $= \Large \frac{\frac{F}{2^6}}{\frac{F}{2^1}}$

$2N-N-1=10 \\N=11.$

$N=11$ which is total no of flipflop present in system,now the mod of this counter will be $2048$ with states transition as $0\rightarrow1....2047\rightarrow 0$

Edit :

Reasoning :

Every flop will reduce frequency by $2$ and cascaded flipflops are likely to depend on previous one .

$\therefore$ let initial frq $=F$

Then : $\Large\frac{\frac{F}{2^N}}{\frac{F}{2^{\frac{N+1}{2}}}}$ $= \Large \frac{\frac{F}{2^6}}{\frac{F}{2^1}}$

$2N-N-1=10 \\N=11.$

$N=11$ which is total no of flipflop present in system,now the mod of this counter will be $2048$ with states transition as $0\rightarrow1....2047\rightarrow 0$