Answer
The acceleration associated with Earth's motion around the Sun
$(T=1 \mathrm{y}=3.156 \times$ $\left.10^{7} \mathrm{s}\right)$ is
$$
a=g \times \frac{4 \pi^{2}\left(1.5 \times 10^{11} \mathrm{m}\right)}{\left(3.156 \times 10^{7} \mathrm{s}\right)^{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)}=6.1 \times 10^{-4} \mathrm{g}
$$
Work Step by Step
The acceleration associated with Earth's motion around the Sun
$(T=1 \mathrm{y}=3.156 \times$ $\left.10^{7} \mathrm{s}\right)$ is
$$
a=g \times \frac{4 \pi^{2}\left(1.5 \times 10^{11} \mathrm{m}\right)}{\left(3.156 \times 10^{7} \mathrm{s}\right)^{2}\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)}=6.1 \times 10^{-4} \mathrm{g}
$$
