This question came up in a recent research we hold with Lubotzky, Sauer and Weinberger. I will share our findings.

Claim: For every cocompact lattice $\Gamma$ in $G=\text{SL}_3(\mathbb{R})$ there exists a finite index lattice $\Gamma_1\subseteq\Gamma$ with $b_2(\Gamma_1)\neq 0$ (thus also $b_2(\Gamma_2)\neq 0$ for every finite index $\Gamma_2\subseteq \Gamma_1$).

Proof: We assume without loss of the generality that $\Gamma$ is torsion free. By [1, Theorem B], there exists a finite index subgroup $\Gamma_0\subseteq \Gamma$ which surjects on $\mathbb{Z}/2\times \mathbb{Z}/2$. We let $\Gamma_1\lhd \Gamma_0$ be the kernel of this surjection and consider $M=\Gamma_1\backslash G/K$. Note that $\Gamma_0/\Gamma_1$ acts on $M$. By [2, Theorem D] (due to Weinberger and Davis) we get that $M$ is not a rational homology sphere. Since $\Gamma_0$ is a Poincare duality group of dimension 5 and $b_1(\Gamma)=0$ (by property T), we conclude that $b_2(\Gamma_1)\neq 0$. $\square$

The claim also hold for nonuniform lattices by a different method.

[1]: Lubotzky, Alexander On finite index subgroups of linear groups. Bull. London Math. Soc. 19 (1987), no. 4, 325–328.

[2]: Davis, James F. The surgery semicharacteristic. Proc. London Math. Soc. (3) 47 (1983), no. 3, 411–428.

This question came up in a recent research we hold with Lubotzky, Sauer and Weinberger. I will share our findings.

Claim: For every cocompact lattice $\Gamma$ in $G=\text{SL}_3(\mathbb{R})$ there exists a finite index lattice $\Gamma_1\subseteq\Gamma$ with $b_2(\Gamma_1)\neq 0$ (thus also $b_2(\Gamma_2)\neq 0$ for every finite index $\Gamma_2\subseteq \Gamma_1$).

Proof: We assume without loss of the generality that $\Gamma$ is torsion free. By [1, Theorem B], there exists a finite index subgroup $\Gamma_0\subseteq \Gamma$ which surjects on $\mathbb{Z}/2\times \mathbb{Z}/2$. We let $\Gamma_1\lhd \Gamma_0$ be the kernel of this surjection and consider $M=\Gamma_1\backslash G/K$. Note that $\Gamma_0/\Gamma_1$ acts on $M$. By [2, Theorem D] (due to Davis and Weinberger) we get that $M$ is not a rational homology sphere. Since $\Gamma_0$ is a Poincare duality group of dimension 5 and $b_1(\Gamma)=0$ (by property T), we conclude that $b_2(\Gamma_1)\neq 0$. $\square$

The claim also holds for nonuniform lattices by a different method.

[1]: Lubotzky, Alexander On finite index subgroups of linear groups. Bull. London Math. Soc. 19 (1987), no. 4, 325–328.

[2]: Davis, James F. The surgery semicharacteristic. Proc. London Math. Soc. (3) 47 (1983), no. 3, 411–428.