The unit of the rate constant $k$ is given as $\text{sec}^{-1}$, which indicates that the decomposition of $N_2O_5$ is a first-order reaction.

For a first-order reaction, the rate law is expressed as: $\text{Rate} = k[N_2O_5]$

Given: $\text{Rate} = 1.02 \times 10^{-4} \text{ mol L}^{-1} \text{ sec}^{-1}$ Rate constant, $k = 3.4 \times 10^{-5} \text{ sec}^{-1}$

Rearranging the rate law to solve for the concentration of $N_2O_5$: $[N_2O_5] = \frac{\text{Rate}}{k}$ $[N_2O_5] = \frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}} = \frac{10.2 \times 10^{-5}}{3.4 \times 10^{-5}} = 3.0 \text{ mol L}^{-1}$