A positive integer $n$ can be described as $B$-rough if all of the prime factors of $n$ strictly exceed $B$.
The first five 2-rough numbers are 1, 3, 5, 7, 9. Note that we always include 1 by convention.
It appears to be true from numerical testing that the $k$th $B$-rough number will never exceed $Bk$.
How could one prove this?
The claim is clear for $B\le 2$, but also for $B\le 4$ as we then count all numbers $\equiv \pm1\pmod 6$. With a few of manual checks (and considering residue classes modulo $30$), we also treat the cases $B=5$ and $B=6$. Even more manualchecking solves the case $B=7$ (and at the same time $B=8,9,10$).
The claim is also clear for $k=1$ as $1$ is always $B$-rough, and for $k=2$ because there is a prime between $B$ and $2B$ by Bertrand's postulate.
We find some explicit bounds for the prime-counting function, e.g., $$\frac x{\ln x}<\pi(x)<1.25506\frac x{\ln x}\qquad \text{for }x\ge 17$$ (with the upper bound already holding for $x>1$).
This makes $$\tag1\pi(kB)-\pi(B)>\frac{kB}{\ln{kB}}-1.25506\frac B{\ln B}>\frac{(k-1.25506)B}{\ln kB}$$ for $k\ge 3$ and $B\ge 7$. By verifying that this is $>k-2$ for $B=11$ and $k=3,4,5,\ldots, 5451$, we solve the case for $k\le 5451$ and arbitrary $B$.
For $B=11$, the set of $B$-rough numbers is periodic modulo $11\#=2310$, hence already the correctness for all $k\le 210$ solves the case $B=11$ for arbitrary $k$. Likewise, the correctness for all $k\le 2310$ solves the case $B=13$ for arbitrary $k$. So we continue comparing $(1)$ to $k-2$, but now with $B=17$. This allows us to go up until $k=1420893$. Again, we need only the cases up to $k\le 30030$ to solve $B=17$ for all $k$, and up to $k\le 510510$ to solve $B=19$ for all $k$.
In the light of Ross Millikjan's answer, we have shown that we "will not fail for $k$ small" as long as we take small to mean $\le 1420893$.
It is time to use sharper bounds, such as Dusart[2010], $$ \frac{x}{\ln x-1}<pi(x)<\frac x{\ln x-1.1}\qquad\text{for }x\ge 60184.$$ For $B\ge 23$, $k_0B>60184$ with $k_0=2617$. Together with the previous results, this makes (for $k>1420893$) $$ \begin{align}\pi(kB)-\pi(B)&=\pi(kB)-\pi(k_0B)+\pi(k_0B)-\pi(B)\\ &>\frac{kB}{\ln kB-1}-\frac{k_0B}{\ln k_0B-1.1}+k_0-1\\ &>\frac{(k-k_0)B}{\ln kB-1}+k_0-1\end{align}$$
Suffice it to say that this takes us to a lot larger values of $k$, we can increase $B$ again, etc. However, I do not know if this method will eliminate the "fail for small $k$" problem for all $B$ ...
The second $B-$rough number is the first prime above $B$. Bertrand's postulate tells us there is a prime between $B$ and $2B$, so you will never fail for $k=2$. For $B=5$ the $B-$rough numbers are those equivalent to $1,7,11,13,17,19,23,29 \bmod 30$, which is $8$ of them in every $30$ so once we do not fail for $k \le 6$ we will never fail.
If we were to fail for some $B$ we would also fail for the next prime below $B$, so we only need to check prime values of $B$. We can show by induction that we will never fail in the long term. The modulus of interest is $B\#$, the product of the primes up to $B$, the second definition of primorial in Wikipedia. We want to make sure there are at least $\frac {B\#}B$ residues up to $B\#$ that are coprime to all the primes less than or equal to $B$. If $A$ is the prime below $B$ and there were sufficient residues at $A$, we now have $B-1$ times as many residues and the required number is only multiplied by $A$, so there will be enough. I haven't found how to justify that enough of these residues will be small that we will not fail for $k$ small.
S. Fan, in Theorem 2.3 of https://www.sciencedirect.com/science/article/pii/S0022314X24000349, proves that
$$\Phi(x,B)>.4\,x/\log B$$
uniformly for $7\le B\le x^{2/3}$ (read the paper for details; note the author uses $y$ instead of $B$). Here, $\Phi(x,B)$ is the number of $B$-rough numbers less or equal than $x$. From this you should be able to derive the result you are after. To illustrate, if $x$ were 100, say, and $B$ were 7, you have at least 40/2=20 rough numbers below 100; considering that 20*7=140, you would know that the 20-th $B$-rough number is below 100, therefore below 140, as you are suggesting.
As TRJ points out, “more importantly deduce from Fan that in general there are more than $.4\,x/\log B$ numbers less than $x$ that are $B$-rough (under the assumptions of the theorem). So if you have $k<.4\,x/\log B$, then the $k$-th $B$-rough number is less than $x$. Equivalently, the $k$-th $B$-rough number is less than $x$ already when $x > 2.5 k\log B$, which is a lot smaller than $kB$, when $B$ is not very small.”
Fan (and Pomerance) have several results concerning rough numbers which you might find interesting, just follow the articles in the bibliography of the linked article. All of them rather recent.
In a different article by Schroeder (not peer reviewed that I know, see arXiv article number 1705.04831v2), you have
$$\Phi(x,B)>\lfloor \frac{2x}{B}\rfloor +1.$$
In this case, $B$ is assumed to be prime which is at least 11. Again, letting $x=kB$, you end up with at least $2k+1$ numbers that are $B$-rough below $kB$, so your assertion follows.
