The question has been answered before. However, when I looked at it, I couldn't get it. So I am providing an alternate explanation here.

By the basic performance equation, we know that $$ \text{Execution Time} = \frac{N \times S}{ R} $$ where N is the number of instructions, S is the average clock cycles taken per instruction and R is the number of clock cycles per second.

Another useful metric is throughput, which is the number of instructions executed per second. It is defined as:

$$\text{Throughput} = \frac{R}{S}$$.

Now in an non-pipelined processor, the number of stages is equal to the number of cycles i.e $S = \text{No. of stages}$

In a pipelined processor, the value of $S = 1$ because an instruction can enter the pipeline in every cycle.

However, this is the case for an ideal pipeline. In case of a pipeline with stall, the general formula is:

$$S = 1 + \delta_{stall} + \delta_{branch} + \delta_{cache}$$ where $\delta$ is the penalty associated with each of the misses.

Here, $\delta_{stall}$ is given as $0.25 \times 2$, hence the value of $S$ is $1+ 0.5 = 1.5$.

So according to the question, non-pipelined $T_{np} = R/6$ and for the pipelined, $T_{p} = R/1.5$.

From this, we can see that the speedup is $4$ times.