Answer
a) $45.9048\;\rm eV$
b) $17.3261 \;\rm eV$
Work Step by Step
a) In this part, the author asks us to assume that the mass of the nucleons is 2$\%$ of the critical mass which means that the mass of neutrinos is the r 98$\%$ of the critical mass density.
So that,
$$\rho_{\rm nucleon}=0.02\rho_c\tag 1 $$
where $\rho_c$ is the critical density of the universe.
$$\;\sigma_{\rm nucleon}=\dfrac{\rho_{\rm nucleon}}{\rho_{\rm neutrino}}=\dfrac{0.02\rho_c}{\rho_{\rm proton}}\tag 2$$
$\star$ $\sigma$ is the number density.
We put the mass density of the proton since the author told us that the mass density of neutrino is assumed to be the mass density of the proton.
$$\sigma_{\rm neutrino}=10^9 N_{\rm nucleon}$$
Plugging from (2)
$$\sigma_{\rm neutrino}=10^9\times \dfrac{0.02\rho_c}{\rho_{\rm proton}} $$
Plugging the known;
$$\sigma_{\rm neutrino}=10^9\times \dfrac{0.02\times 10^{-26}}{1.67\times 10^{-27}}=\bf 1.1976\times 10^8 \;\rm \nu/m^3$$
So, the mass of the neutrino is given by
$$m_\nu=0.98\times \dfrac{\rho_c}{\sigma_{\rm neutrino}}$$
Plugging the known;
$$m_\nu=0.98\times \dfrac{10^{-26}}{ 1.1976\times 10^8}=\bf 8.18303\times 10^{-35}\;\rm kg$$
Now we need to convert the mass to eV.
$$m_\nu=\rm 8.18303\times 10^{-35}\;\rm \color{red}{\bf\not}kg\times \dfrac{931.5\times 10^6\;eV}{1.6605\times 10^{-27}\;\color{red}{\bf\not}kg}$$
$$m_\nu=\color{red}{\bf 45.9048}\;\rm eV $$
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b) In this part, the author asks us to assume that the mass of the nucleons is 5$\%$ of the critical mass which means that the mass of neutrinos is the r 95$\%$ of the critical mass density.
By the same approach as we did in the first part;
$$\sigma_{\rm neutrino}=10^9\times \dfrac{0.05\times 10^{-26}}{1.67\times 10^{-27}}=\bf 2.99401\times 10^8 \;\rm \nu/m^3$$
So, the mass of the neutrino is given by
$$m_\nu=0.95\times \dfrac{\rho_c}{\sigma_{\rm neutrino}}$$
Plugging the known;
$$m_\nu=0.95\times \dfrac{10^{-26}}{ 2.99401\times 10^8}=\bf 3.173\times 10^{-35}\;\rm kg$$
Now we need to convert the mass to eV.
$$m_\nu=\rm 3.173\times 10^{-35}\;\rm \color{red}{\bf\not}kg\times \dfrac{931.5\times 10^6\;eV}{1.6605\times 10^{-27}\;\color{red}{\bf\not}kg}$$
$$m_\nu=\color{red}{\bf 17.3261}\;\rm eV $$
