For radioactive decay, $N(t)=N_0 e^{-t/\tau_m}$ where $\tau_m$ is the mean life.

Activity is $A=-\frac{dN}{dt}=\frac{N}{\tau_m}$ so activity also decays exponentially: $A(t)=A_0 e^{-t/\tau_m}$

The rate of change of activity is defined as $R=-\frac{dA}{dt}$ Since $A(t)=A_0 e^{-t/\tau_m},$ we get $\frac{dA}{dt}=-\frac{A_0}{\tau_m}e^{-t/\tau_m}=-\frac{A}{\tau_m}$ Hence, $R=\frac{A}{\tau_m}$

Now treat the two sources separately.

Source P: mean life $\tau$ $A_P(t)=A_{P0}e^{-t/\tau}$ $R_P(t)=\frac{A_P(t)}{\tau}=\frac{A_{P0}}{\tau}e^{-t/\tau}$

At $t=2\tau$, $R_P=\frac{A_{P0}}{\tau}e^{-2}$

Source Q: mean life $2\tau$ $A_Q(t)=A_{Q0}e^{-t/(2\tau)}$ $R_Q(t)=\frac{A_Q(t)}{2\tau}=\frac{A_{Q0}}{2\tau}e^{-t/(2\tau)}$

At $t=2\tau$, $R_Q=\frac{A_{Q0}}{2\tau}e^{-1}$

Given that both sources have the same activity at $t=0$, $A_{P0}=A_{Q0}$

Therefore, \frac{R_P}{R_Q}=rac{\frac{A_{P0}}{\tau}e^{-2}}{\frac{A_{Q0}}{2\tau}e^{-1}} Using $A_{P0}=A_{Q0}$, $\frac{R_P}{R_Q}=\frac{1}{\tau}e^{-2}\cdot \frac{2\tau}{1}e^{1}=2e^{-1}=\frac{2}{e}$

Comparing with $\frac{R_P}{R_Q}=\frac{n}{e}$ we get $n=2$