Too big to compute

$\begin{aligned} \color{#3D99F6}{6911}^{6990^{9613}} & \equiv 6911^{\color{#3D99F6}{6990^{9613} \text{ mod } \phi (100)}} \text{ (mod }\color{#3D99F6}{100}) \quad \quad \small \color{#3D99F6}{\text{As 6911 and 100 are coprimes, we can apply Euler's theorem.}} \\ & \equiv 6911^{6990^{9613} \text{ mod } 40} \text{ (mod 100)} \quad \quad \small \color{#3D99F6}{\text{Euler's totient function } \phi(100) = 40} \\ & \equiv 6911^{\color{#3D99F6}{30^{9613} \text{ mod } 40}} \text{ (mod 100)} \quad \quad \small \color{#3D99F6}{6990 = 174\times 40 + 30} \\ & \equiv 6911^{\color{#3D99F6}{0}} \text{ (mod 100)} \quad \quad \small \color{#3D99F6}{30^n \equiv 0 \text{ (mod 40) for } n \ge 3} \\ & \equiv \boxed{1} \text{ (mod 100)} \end{aligned}$

$6911^{6990^{9613}}\equiv 11^{6990^{9613}}\pmod{100}\\ 11^{\lambda(100)}=11^{20}\equiv 1\pmod{100}\\ \implies \boxed{6990^{9613}\pmod{20}}\\ 6990^{9613}\equiv 0\pmod{20}\\ \implies 11^{6990^{9613}}\equiv 11^0\equiv \boxed{01}\pmod{100}$

Clearly, $6911\equiv 11(mod\quad 100)$ , so now we are working with the powers of $11$ . Let $n\equiv k(mod\quad 10)$ , and note that ${ 11 }^{ n }\equiv 10k+1(mod\quad 100)$ . As $6990\equiv 0(mod\quad 10)$ , then ${ 11 }^{ 6990 }\equiv 1(mod\quad 100)$ , therefore, $\ 6911^{6990^{9613}}\equiv 1(mod\quad 100)$