, 2021

Solution:
Notice that this value, in base $b$, is
$$\frac{b^{6}-1}{b-1} = (b+1)\left(b^{2}-b+1\right)\left(b^{2}+b+1\right)$$
This means that, if $b$ satisfies the problem condition, $(b+1)\left(b^{2}-b+1\right)\left(b^{2}+b+1\right)$ has more than $p_{1} \ldots p_{b}$, where $p_{i}$ is the $i$th smallest prime.
We claim that, if $b \geq 7$, then $p_{1} \ldots p_{b} > (b+1)\left(b^{2}-b+1\right)\left(b^{2}+b+1\right)$. This is true for $b=7$ by calculation, and can be proven for larger $b$ by induction and the estimate $p_{i} \geq i$.
All we have to do is to check $b \in 2,3,4,5,6$. Notice that for $b=6$, the primes cannot include $2,3$ and hence we want $\frac{6^{6}-1}{5}$ to be divisible by a product of 6 primes the smallest of which is 5. However, $5 \cdot 7 \cdots 17 > \frac{6^{6}-1}{5}$, and by checking we rule out 5 too. All that is left is $\{2,3,4\}$, all of which work, giving us an answer of 24.