Consider a 3-bit ripple counter. It can count from 0 to 7.
The count will be: 000, 001, 010, 011, 100, 101, 110, 111.
In order to go from 3 to 4, all the bits have to be flipped. Since this is ripple (asynchronous) counter — it lacks a global clock and hence all the bits would not change simultaneously.
$Q_0$ will change first, then $Q_1$, then $Q_2$.
This will encounter the worst case delay, and hence for ripple counters to function properly, the clock period to be used must be $\geq$ combined propagation delays of the constituent Flip Flops.
Therefore, for an n-bit Ripple Counter
$T_{clk} \geq n*T_p$
If we factor in the synchronisation signal (ie strobe signal), then
$T_{clk} \geq n*T_p+T_{strobe}$
Now, take the minimum case.
$T_{clk} = 8*75+50=650ns$
If $T=650ns$
=> $F=\frac{1}{650*10^{-9}}Hz$
=> $F = 1.5*10^6Hz$
Option A