$$ A = \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2^2\mathbb{Z} \oplus \mathbb{Z}/3^2\mathbb{Z} \oplus \mathbb{Z}/3^3\mathbb{Z} \oplus \mathbb{Z}/5\mathbb{Z} \oplus \mathbb{Z}/5^2\mathbb{Z} \oplus \mathbb{Z}/5^2\mathbb{Z} \oplus \mathbb{Z}/5^2\mathbb{Z} $$
First of all, $A$ is isomorphic to an abelian group of order $121500000$. Then, it can be split to an isomorphic group of $\mathbb{Z}/64\mathbb{Z} \oplus \mathbb{Z}/243\mathbb{Z} \oplus \mathbb{Z}/78125\mathbb{Z}$. But how should I proceed next?
There is a fact that serves to find them algorithmically.
Fact. Let $G$ be a finitely generated group, then there exists $m\geq 0$, $p_1,\ldots,p_n\in \mathbb{Z}_{+}$, distinct primes and naturals \begin{align} &\alpha_{11}\geq\cdots\geq\alpha_{1k_{1}}>0\\ &\vdots\\ &\alpha_{n1}\geq\cdots\geq \alpha_{nk_{n}}>0 \end{align} such that $$G\cong \mathbb{Z}^{(m)}\oplus\mathbb{Z}_{p_1^{\alpha_{11}}}\oplus\cdots\oplus\mathbb{Z}_{p_{1}^{\alpha_{1k_1}}}\oplus\cdots\oplus\mathbb{Z}_{p_{n}^{\alpha_{n1}}}\oplus\cdots \oplus\mathbb{Z}_{p_{n}^{\alpha_{nk_n}}}$$ Let $q_{i}=p_{1}^{\alpha_{1i}}p_{2}^{\alpha_{2i}}\cdots p_{n}^{\alpha_{ni}}$, these elements are called invariant factors, and satisfies that $q_{i}\mid q_{i-1}$ and $$G\cong \mathbb{Z}^{(m)}\oplus \mathbb{Z}_{q_1}\oplus\cdots \mathbb{Z}_{q_n}$$
In your case $|G|=2^{6}\times 3^{5}\times 5^{7}$, computing the proper partitions you can verify that there are $1155$ finitely generated groups except for isomorphisms, in particular:
\begin{equation} A\cong \mathbb{Z}_{2^2}\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\oplus\mathbb{Z}_{3^3}\oplus\mathbb{Z}_{3^2}\oplus\mathbb{Z}_{5^2}\oplus\mathbb{Z}_{5^2}\oplus\mathbb{Z}_{5} \end{equation}
\begin{align} p_{1}&=2:= 2 \quad 1 \quad 1 \quad 1 \quad 1\\ p_{2}&=3:= 3 \quad 2 \quad 0 \quad 0 \quad 0\\ p_{3}&=5:= 2 \quad 2 \quad 2 \quad 1 \quad 0 \end{align} then, the invariant factors are \begin{align} q_{1}&=2^3\times 3^3\times 5^2\\ q_{2}&=2^{1}\times 3^{2}\times 5^{2}\\ q_{3}&=2^{1}\times 5^{2}\\ q_{4}&=2\times 5\\ q_{5}&=2\\ \end{align} Hence \begin{equation} A\cong \mathbb{Z}_{q_{1}}\oplus\mathbb{Z}_{q_{2}}\oplus\mathbb{Z}_{q_{3}}\oplus\mathbb{Z}_{q_{4}}\oplus\mathbb{Z}_{q_{5}}. \end{equation}