The formula can be derived the following way:
We are dealing with numbers of the form
$$ \hspace{120pt} z = z_1,z_2\dots z_p \cdot 10^m \hspace{100pt} (1)$$
where $z_1,z_p \in \{1,\dots,9\}$ and $z_i \in \{0,\dots,9\}$ for $1<i<p$, so $z$ is in the interval $(10^m, 10^{m+1})$. Moreover, there is an integer $n$ such that $z\in[2^n, 2^{n+1})$.
When we store the number in a computer, it is converted to binary floating-point format, that is, to a number
$$ \hspace{120pt} y = y_1,y_2\dots y_q \cdot 2^n \hspace{100pt} (2)$$
close to $z$, where $y_1=1$ and $y_i \in \{0,1\}$ for $i>1$. Since the $n$ in this formula is the specific $n$ from above, we need to consider that this $y$ can also be $1\cdot 2^{n+1}$. Since $y$ is closer to $z$ than any other such number in $[2^n,2^{n+1}]$, we follow that $y$ and $z$ have at most half the distance between these numbers:
$$ |y-z| \le \tfrac12 2^{-(q-1)}2^n $$
Now, in order to deduce that the stored number $y$ must have come from the number $z$, this $z$ must lie closer to $y$ than any other number of that form, that is
$$ |y-z| < \tfrac12 10^{-(p-1)}10^m $$
We would like to have this as a sufficient condition.
The case $z=10^m$ is special, as $z$ could be rounded down to a $y<z$. That would mean $y\in(10^{m-1},10^m]$, where it would have to satisfy $|y-z| < \tfrac12 10^{-(p-1)}10^{m-1}$. It is therefore appropriate to consider $z$ as an element of $(10^{m-1},10^m]$.
We conclude that $z$ can be stored exactly if
$$ \tfrac12 2^{-(q-1)}2^n < \tfrac12 10^{-(p-1)}10^m $$
or, equivalently,
$$ p < (q-1-n)\log_{10}2 + (m+1) $$
To see how $n$ and $m$ play into this, imagine the real line divided by powers of $2$ resp. $10$ into segments of either of the forms $(10^m, 2^n]$, $(2^n, 2^{n+1}]$, $(2^n, 10^m]$. For example, the interval $(10^3,10^4]$ consists of five segments:
$$ (10^3,2^{10}],\ (2^{10},2^{11}],\ (2^{11},2^{12}],\ (2^{12},2^{13}],\ (2^{13},10^4] $$
In each segment, $m$ and $n$ have a certain value, for example, in the first segment above, $m=3$ and $n=9$, while in the last segment, $m=3$ and $n=13$. When $q$ is a fixed number, the numbers of the form $(2)$ divide an interval $(2^n,2^{n+1}]$ into $2^{q-1}$ steps. The larger $q$ and the smaller $n$, the smaller the steps are, and the better can $z$ be approximated. However, the larger $p$ is, the closer the approximation $y$ lies to a value different from $z$. This is reflected by the condition above being easier to satisfy when $p$ is small, $q$ is large, and $n$ is small. $p$ must be the smallest for numbers in an interval of the form $(2^n,10^{m+1}]$.
Let's fix $q=24$. The condition above can be written as
$$ p < 23\log_{10}2 - n\log_{10}2 + (m+1) $$
Since $n$ can only have values such that $\tfrac12 10^m < 2^n < 10^{m+1}$, we see that the middle summand satisfies
$$ m - \log_{10}2 < n\log_{10}2 < m+1 $$
Thus, the right-hand side takes values in
$$ (23\log_{10}2,\ \ 23\log_{10}2 + \log_{10}2 + 1) $$
Since $\log_{10}2$ is irrational, $n\log_{10}2$ frequently comes arbitrarily close to $m+1$, so these values come close to $23\log_{10}2 \approx 6.92$. That means every now and then, $p$ must be as low as $6$. This is the case, for example, when $m=9$ and $n=33$, where $p$ must be smaller than $6.9897$.
