Answer
e. $<\frac{1}{2}$
Work Step by Step
The lifetime of an excited state is given by:
$$\tau = \frac{1}{r}$$
where $r$ is the decay rate.
If the decay rate of atom $B$ is is twice that of atom $A$ then,
$$r_B = 2r_A$$
therefore, $$\tau_A = \frac{2}{r_B}$$
and, $$\tau_B = \frac{1}{r_B}$$
The decay equation can be expressed as:
$$N_{exc} = N_0 e^{-t/\tau}$$
Taking the ration of $\frac{N_B}{N_A}$ gives:
$$\frac{N_B}{N_A} = \frac{N_0 e^{-t/\tau_B}}{N_0 e^{-t/\tau_A}}$$
Now plugging in $\tau_A = \frac{2}{r_B}$, $\tau_B = \frac{1}{r_B}$, $t = \tau_A$ as stated in the problem, and canceling the $N_0$'s
We see that:
$$\frac{N_B}{N_A} = \frac{e^{-2}}{e^{-1}} = e^{-1}$$
Finally, we see that $e^{-1}<\frac{1}{2}$ and the answer is therefore, d.
