Solution:

There are no balanced numbers with $c$ digits if $c > 100$. Indeed, such a number would be the product of $c$ primes, at least half of which are greater than $100$. Hence the number would be greater than $100^{c / 2} = 10^{c}$, which is absurd. In fact the largest balanced number has exactly ten digits, since $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 < 10^{10}$, while $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 > 10^{11}$.