= Solution
Introduce nonterminals for the terminals and one for the pair $bb$:
$$
A\to a,
\qquad B\to b,
\qquad C\to c,
\qquad U\to BB.
$$
Because the original start symbol occurs on a right-hand side, take a fresh start symbol $S_0$. After eliminating the <unit production> $S\to T$, one grammar in <Chomsky normal form> is
$$
\begin{aligned}
S_0&\to SU\mid AS\mid CC,\\
S&\to SU\mid AS\mid CC,\\
U&\to BB,\\
A&\to a,\qquad B\to b,\qquad C\to c.
\end{aligned}
$$
The original grammar generates exactly
$$
\{a^i cc b^{2j}:i,j\geq0\}.
$$
The displayed grammar generates the same set: $AS$ adds an $a$ on the left, $SU$ adds one $bb$ pair on the right, and $CC$ terminates the derivation with $cc$. Since the original grammar does not generate $\epsilon$,
$$
\boxed{\mathcal L(G)=\mathcal L(G_{\rm Chom}).}
$$
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