First use the Chinese remainder theorem:
$$\mathbf Z/990\mathbf Z\simeq \mathbf Z/2\mathbf Z\times\mathbf Z/5\mathbf Z\times\mathbf Z/9\mathbf Z\times\mathbf Z/11\mathbf Z. $$
Now
- $67\equiv1 \mod 2$,
- $67\equiv2 \mod 5$, hence it has order $4$ modulo $5$,
- $67\equiv4 \mod 9$, hence it has order $3$ modulo $9$
- $67\equiv1 \mod 11$.
The order of $67$ modulo $990$ is the l.c.m. of its orders modulo $\;2,5,9$ and $11$: $\;\color{red}{12}$, and
$$67^{26^{42^{23}}}\mod 990=67^{26^{42^{23}}\bmod12}$$
Value of the exponent modulo 12:
First observe $26\equiv 2\mod 12$; it is easy to check that, if $n\ge 2$,
$$2^n\equiv\begin{cases} 4\mod 12&\text{if $n$ is even,}\\8\mod 12&\text{if $n$ is odd.}
\end{cases}$$
We conclude that $26^{42^{23}}\equiv2^{42^{23}}\equiv 4 \mod 12$, and
$$67^{26^{42^{23}}}\equiv 67^4\equiv 661\mod 990.$$