Ta có:

\(\log (\log e) + \log (\ln 10) = \log (\log e.\ln 10) = \log 1 = 0 \Rightarrow \log (\ln 10) = - \log (\log e)\).

Ta có: \(\sqrt {{x^2} + 1} > \sqrt {{x^2}} = \;|x|\; \ge - x\) nên \(\sqrt {{x^2} + 1} + x > 0\quad \forall x \in \mathbb{R}\).

Khi đó \(f(x) = a\ln {\left( {x + \sqrt {{x^2} + 1} } \right)^{2021}} + b{\sin ^3}x + 18 = 2021.a\ln \left( {x + \sqrt {{x^2} + 1} } \right) + b{\sin ^3}x + 18\).

Lại có \(\left( {x + \sqrt {{x^2} + 1} } \right)\left( { - x + \sqrt {{x^2} + 1} } \right) = {x^2} + 1 - {x^2} = 1\) nên

\(\ln \left( {x + \sqrt {{x^2} + 1} } \right) = \ln \frac{1}{{ - x + \sqrt {{x^2} + 1} }} = - \ln \left( { - x + \sqrt {{x^2} + 1} } \right)\)

Với mọi x ta có:

\(\begin{array}{l}f( - x) = 2021.a\ln \left( { - x + \sqrt {{x^2} + 1} } \right) + b{\sin ^3}( - x) + 18\\ = - 2021.a\ln \left( {x + \sqrt {{x^2} + 1} } \right) - b{\sin ^3}x + 18\\ \Rightarrow f( - x) + f(x) = 36.\end{array}\)

Mà \(\log (\ln 10) = - \log (\log e).\)

\( \Rightarrow f(\log (\ln 10)) + f(\log (\log e)) = 36 \Rightarrow f(\log (\ln 10)) = 34\). Chọn A.