Again let $m$ be the mass of the point mass. Then, for $r \leq R$ ,
$\begin{align*}F(r) = \frac{G m M_E \frac{4 \pi r^3}{4 \pi R^3}}{r^2} = \frac{ G m M_E r}{R^3}\end{align*}$
Therefore,
$\begin{align*}\frac{F(R)}{F(R/2)}= \frac{\frac{ G M_E R}{R^3}}{\frac{ G M_E R/2}{R^3}} = \boxed{2}\end{align*}$
Therefore, answer (C) is correct.

Solution to 1992 Problem 5