Solution

Let intervals between emitted pulses (in the frame of the airplane) be $T$. We interpret the phrase "intervals $K$ times longer than the intervals of the emitted pulses" for the reflected pulses received at the airplane to be $T+KT$. Suppose the airplane and the rocket coincide at time $t=0$ when the first pulse is emitted (and immediately reflected and received). When the second pulse is emitted at time $t=T$, the rocket is $vT$ ahead of the airplane, where $v$ is the speed of the rocket relative to the airplane. Thus, the second pulse will be reflected at the rocket at time $t=T+vT/(c-v)$ and received at the airplane at time $t=T+2vT/(c-v)$. Setting $KT=2vT/(c-v)$, we obtain $v=Kc/(2+K)$. Since the speed of the ground relative to the airplane is $-\beta c$, the speed of the rocket relative to the ground is

$$\frac{v-(-\beta c)}{1-v(-\beta c)/c^2}=\frac{[K+(2+K)\beta]c}{(2+K)+K\beta},$$

which becomes $\beta c$ if $K=0$, i.e. intervals between reflected pulses equal intervals between emitted pulses.