= Solution
Let $p:\widetilde M\to M$ be the <universal cover>. Its number of sheets is $|\pi_1(M)|$, which is finite, so <compactness> of $M$ makes $\widetilde M$ <compact>. As a <covering space> of a manifold without boundary, it is also a three-manifold without boundary. It is connected and <simply connected> by the definition of <universal cover>.
A <simply connected> <manifold> is orientable: transport of a local orientation around loops gives the orientation character $\pi_1(\widetilde M)\to\{\pm1\}$, and the trivial <fundamental group> leaves no orientation obstruction. The degree-one <Hurewicz theorem>, equivalently the identification of $H_1$ with the <abelianization> of the <fundamental group>, gives $H_1(\widetilde M;\mathbb Z)=0$. The <universal coefficient theorem for cohomology> then gives
$$
H^1(\widetilde M;\mathbb Z)\cong\operatorname{Hom}(H_1(\widetilde M;\mathbb Z),\mathbb Z)=0,
$$
because the additional term $\operatorname{Ext}(H_0(\widetilde M;\mathbb Z),\mathbb Z)$ vanishes for $H_0\cong\mathbb Z$. Integral <Poincare duality> gives $H_2(\widetilde M;\mathbb Z)\cong H^1(\widetilde M;\mathbb Z)=0$. Connectedness gives $H_0\cong\mathbb Z$, and the orientation <fundamental class> gives $H_3\cong\mathbb Z$; higher groups vanish by dimension. Consequently
$$
\boxed{H_i(\widetilde M;\mathbb Z)\cong\begin{cases}\mathbb Z,&i=0,3,\\0,&i\ne0,3.\end{cases}}
$$
Thus the <universal cover of a closed three-manifold with finite fundamental group> is an <integral homology sphere>. No homeomorphism classification of three-manifolds is needed for this homological conclusion.
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