The universal gravitational constant
$\begin{align*}G = 6.67 \cdot 10^{-11} \;\mathrm{m}^3\mathrm{/}\mathrm{(}\mathrm{kg}\;\mathrm{s}^2)\end{align*}$
is given at the beginning of the exam. So, use
$\begin{align*}g = \frac{MG}{R^2}\end{align*}$
and the known value of $g = 9.8 \;\mathrm{m}\mathrm{/}\mathrm{s}^2$ to estimate $M$ . The result is
$\begin{align*}M \approx \boxed{6 \cdot 10^{24} \mbox{ kg}}\end{align*}$

Solution to 1992 Problem 19