Let the primes $p = 3, q = 23$ and $n = p \times q$.
(i) Determine $\phi (n)$.
(ii) Determine $3^{-1} \pmod { \phi (n)}$.
(i) Since $n=p \times q$, where $p$ and $q$ are prime, we can take advantage of the multiplicity of Euler's totient function.
$$ \begin{align} \phi(n) & = \phi(p \times q) \\ \\ & = \phi(3) \times \phi(23) \\ \\ & = (3-1) \times (23-1) \\ \\ \phi(n) & = 44 \end{align} $$
(ii) Since $\gcd(\phi(n),3) = \gcd(44,3)=1$, we can use Euler's Theorem
$$ 3^{\phi(44)} \equiv 1 \pmod {44} $$
Noting that $\phi(44)=20$, and factorising the LHS gives us
$$ 3 \times 3^{20-1} \equiv 1 \pmod {44}$$
By definition of inverse, we have
$$ 3^{-1} \equiv 3^{19} \pmod {\phi(n)} $$
Using $3^{10} \equiv 59049 \equiv 1 \pmod {44}$, we have $3^{19} \equiv (3^{10})\times 3^9 \equiv 19683 \equiv 15 \pmod {44}$
And so
$$ 3^{-1} \equiv 15 \pmod {44}$$