$a=3^{100}$ has primitive root. thus since $(c,3)=1$, the equation $x^b\equiv c\pmod a$ is solvable iff $c^{\frac{\varphi(a)}{\gcd(\varphi(a),b)}}=c^{96}\equiv 1\pmod a$ and the number of solutions is $\gcd(\varphi(a),b)=54$. for $c=1$ we deduce that $\gcd(\varphi(a),z)=54$. thus $z=2^t\cdot 3^3\cdot k$, where $\gcd(k,6)=1$ and $t\geq1$. now it is obvious that the congruence equation $x^b\equiv c\pmod a$ is solvable iff the congruence equation $x^z\equiv c\pmod a$ be solvable, because $\frac{\varphi(a)}{\gcd(\varphi(a),z)}=96$. so the answer is $z=2^t\cdot 3^3\cdot k$, where $\gcd(k,6)=1$ and $t\geq1$.