The relationship between the magnetic dipole moment and the spin angular momentum vector for the proton and the electron are
$\begin{align*}\vec{\mu}_p &= \frac{g_p e}{2 m_p} \vec{S}_p \\\vec{\mu}_e &= \frac{-e}{m_e} \vec{S}_e\end{align*}$
where $g_p = 5.59$ . The statement that the proton has spin $1/2$ is equivalent to saying that
$\begin{align*}\left|\vec{S}_p \right| = \frac{3 \hbar}{4}\end{align*}$
All electrons also have spin $1/2$ , which means that
$\begin{align*}\left|\vec{S}_e \right| = \frac{3 \hbar}{4}\end{align*}$
The ratio of the magnitudes of the intrinsic magnetic moment of this nucleus to that of an electron is
$\begin{align*}\frac{\displaystyle \left|\frac{g_p e}{2 m_p} \vec{S}_p \right| }{\displaystyle \left|\frac{-e}{m_e}\right|} = ...$
Because the mass of the electron is approximately $2000$ times smaller than that of the proton (and the nucleus will have at least one proton in it), the ratio is much less than 1, and answer (E) is correct.

Solution to 1992 Problem 77