Finding time t in a first order reaction (JEE Adv 2019)
The basic law of chemical kinetics states for $\ce{aR -&gt; bP}$, that
$$A = kR = -\frac{1}{a}\frac{\mathrm dR}{\mathrm dt} = \frac{1}{b}\frac{\mathrm dp}{\mathrm dt}$$
(where $A$ is activity and $k$ is called rate constant1)
Note that, in our case, the stoichiometric coefficient of reactant (i.e., $\ce{N2O5}$) is 2, therefore the differential rate equation becomes,
$$-\frac{1}{2}\frac{\mathrm dR}{\mathrm dt} = kR$$
or,
$$-\frac{\mathrm dR}{\mathrm dt} = (2k)R$$
On integrating and re-arranging, we get,
$$R = R_0 \mathrm e^{(2k)t}$$
which means, $$t = \frac{2.303}{2k}\log_{10}{\frac{R_0}{R}}$$
This factor of $\frac{1}{2}$ was what you missed, I think.

1: As per my first equation, it's clear how rate constant is specific to a certain reaction. The following is an extract from Concentration and Rates (Differential Rate Laws) (under the heading &quot;Reaction Orders&quot;):

The proportionality constant (k) is called the rate constant, and its value is characteristic of the reaction and the reaction conditions. A given reaction has a particular rate constant value under a given set of conditions, such as temperature, pressure, and solvent.