Answer
$$
\frac{m}{4 \pi r^{3} / 3}=\frac{M_{E}}{4 \pi R_{E}^{3} / 3} \Rightarrow m=M_{E}\left(\frac{r}{R_{E}}\right)^{3}
$$
$\left.\text { which yields (with } M_{E} \approx 6 \times 10^{24} \mathrm{kg} \text { and } R_{E} \approx 6.4 \times 10^{6} \mathrm{m}\right) m=2.3 \times 10^{7} \mathrm{kg}$
With the above assumptions, the acceleration due to gravity is
$$
a_{g}=\frac{G m}{r^{2}}=\frac{\left(6.7 \times 10^{-11} \mathrm{m}^{3} / \mathrm{s}^{2} \cdot \mathrm{kg}\right)\left(2.3 \times 10^{7} \mathrm{kg}\right)}{(10 \mathrm{m})^{2}}$$$$=1.5 \times 10^{-5} \mathrm{m} / \mathrm{s}^{2} \approx 2 \times 10^{-5} \mathrm{m} / \mathrm{s}^{2}
$$
Work Step by Step
$$
\frac{m}{4 \pi r^{3} / 3}=\frac{M_{E}}{4 \pi R_{E}^{3} / 3} \Rightarrow m=M_{E}\left(\frac{r}{R_{E}}\right)^{3}
$$
$\left.\text { which yields (with } M_{E} \approx 6 \times 10^{24} \mathrm{kg} \text { and } R_{E} \approx 6.4 \times 10^{6} \mathrm{m}\right) m=2.3 \times 10^{7} \mathrm{kg}$
With the above assumptions, the acceleration due to gravity is
$$
a_{g}=\frac{G m}{r^{2}}=\frac{\left(6.7 \times 10^{-11} \mathrm{m}^{3} / \mathrm{s}^{2} \cdot \mathrm{kg}\right)\left(2.3 \times 10^{7} \mathrm{kg}\right)}{(10 \mathrm{m})^{2}}$$$$=1.5 \times 10^{-5} \mathrm{m} / \mathrm{s}^{2} \approx 2 \times 10^{-5} \mathrm{m} / \mathrm{s}^{2}
$$
