Solution:

Any even number $2n$ can be written as $22$ in base $n - 1$. In particular, $1994 = 22_{996}$.

We have to show that we cannot write $1993 = aaa\ldots a_b$. If the number has $n$ digits, then
$$1993 = a(1 + b + \ldots + b^{n-1}) = a\frac{b^n - 1}{b - 1}.$$
But $1993$ is prime, so $a$ must be $1$. Hence
$$b^{n-1} + \ldots + b - 1992 = 0.$$
So $b$ must divide $1992 = 2^3 \cdot 3 \cdot 83$. We cannot have $n = 2$, for then $b = 1992$ and we require $b < 1992$. So $n > 2$. But $83^2 = 6889 > 1993$, so $b$ must divide $24$. Hence $b = 2, 3, 4, 6, 8, 12$, or $24$. But we can easily check that none of these work:

$1 + 2 + 2^2 + \ldots + 2^9 = 1023$, $1 + \ldots + 2^{10} = 2047$.

$1 + 3 + \ldots + 3^6 = 1093$, $1 + \ldots + 3^7 = 3280$

$1 + 4 + \ldots + 4^5 = 1365$, $1 + \ldots + 4^6 = 5461$

$1 + 6 + \ldots + 6^4 = 1555$, $1 + \ldots + 6^5 = 9331$

$1 + 8 + 8^2 + 8^3 = 585$, $1 + \ldots + 8^4 = 4681$

$1 + 12 + 12^2 + 12^3 = 1885$, $1 + \ldots + 12^4 = 22621$

$1 + 24 + 24^2 = 601$, $1 + \ldots + 24^3 = 14425$