Answer
(a)
$$ \approx 2.1 \times 10^{-26} \mathrm{kg} / \mathrm{m}^{3}$$
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(b)
$$\rho_{\mathrm{dark}} \approx \left|1.0 \times 10^{-25} \mathrm{kg} / \mathrm{m}^{3}\right|$$
Work Step by Step
(a)
note that The data in the Problem are for visible matter only (stars and galaxies).
$$\rho_{\text {baryon }}=10 \times \rho_{\text {visible }}=10 \times \frac{M_{\text {visible }}}{\frac{4}{3} \times \pi \times R^{3}}$$
$$=10 \frac{\left(10^{11} \text { galaxies }\right)\left(10^{11} \text { stars/galaxy }\right)\left(2.0 \times 10^{30} \mathrm{kg} / \mathrm{star}\right)}{\frac{4}{3} \pi\left[\left(14 \times 10^{9} \mathrm{ly}\right)\left(9.46 \times 10^{15} \mathrm{m} / \mathrm{ly}\right)\right]^{3}} \approx $$
$$ \approx 2.1 \times 10^{-26} \mathrm{kg} / \mathrm{m}^{3}$$
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(b)
Again, according to the text, dark matter is about 5 times more plentiful than baryonic matter
$$\rho_{\mathrm{dark}}=5\times \rho_{\mathrm{baryon}}=5 \times \left(2.055 \times 10^{-26} \mathrm{kg} / \mathrm{m}^{3}\right) \approx\left|1.0 \times 10^{-25} \mathrm{kg} / \mathrm{m}^{3}\right|$$
