Answer
The acceleration associated with the Solar System's motion around the galactic center
$\left(T=2.5 \times 10^{8} \mathrm{y}=7.9 \times 10^{15} \mathrm{s}\right)$ is
$$a=g \frac{4 \pi^{2}\left(2.2 \times 10^{20} \mathrm{m}\right)}{\left(7.9 \times 10^{15} \mathrm{s}\right)^{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)}$$$$=1.4 \times 10^{-11} \mathrm{g}$$
Work Step by Step
The acceleration associated with the Solar System's motion around the galactic center
$\left(T=2.5 \times 10^{8} \mathrm{y}=7.9 \times 10^{15} \mathrm{s}\right)$ is
$$a=g \frac{4 \pi^{2}\left(2.2 \times 10^{20} \mathrm{m}\right)}{\left(7.9 \times 10^{15} \mathrm{s}\right)^{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)}$$$$=1.4 \times 10^{-11} \mathrm{g}$$
