Why isn't the answer working?

Fermat's Little Theorem states that if $p$ is prime and $\gcd{a, p} = 1$ , then

$a^{p-1} \equiv 1 \pmod{p}$

Using this theorem, we get this:

$2^{10} \equiv 1 \pmod{11}$

Which further leads to

$2^{1993} \equiv 2^3 \equiv 8 \pmod{11}$

Hence, $8$ is the smallest answer. But the question asks for the second smallest so we add $11$ to the answer to get the actual answer as $19$ .

i did it with simple congruence ${ 2 }^{ 5\quad }\equiv \quad -1(mod\quad 11)\\ 2^{ 1990 }\quad \equiv \quad 1(mod\quad 11)\\ { 2 }^{ 3 }\quad \equiv \quad -3(mod\quad 11)\\ { 2 }^{ 1993 }\quad \equiv \quad -3(mod\quad 11)\\ { 2 }^{ 1993 }\quad \equiv \quad 8\quad (mod\quad 11)\\ 8\quad \equiv \quad 19\quad (mod\quad 11)\\ { 2 }^{ 1993 }\quad \equiv \quad 19\quad (mod\quad 11)\\ \quad$