From Mechanics

Consider a particle on the surface of the planet.

\displaystyle F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{\, 2}}}

where \displaystyle F is the gravitational force on a particle on the surface of the planet, in other words, its weight.

\displaystyle {{m}_{1}} is the mass of the planet, \displaystyle {{m}_{2}} is the mass of the particle and \displaystyle d is the radius of the planet.

As \displaystyle F={{m}_{2}}a where \displaystyle a is the acceleration of the particle very close to the surface of the planet,

\displaystyle a=\frac{G{{m}_{1}}}{{{d}^{\, 2}}}

Thus

\displaystyle {{d}^{\, 2}}=\frac{G{{m}_{1}}}{a}

As \displaystyle G=6\textrm{.}67\times {{10}^{-11}}\text{ k}{{\text{g}}^{\text{-1}}}{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{-2}}} we get

\displaystyle \begin{align} & {{d}^{\, 2}}=\frac{\left( 6\textrm{.}67\times {{10}^{-11}} \right)\times \left( 5\times {{10}^{20}} \right)}{{{3\textrm{.}2}^{{}}}}=\frac{33\textrm{.}35\times {{10}^{9}}}{3\textrm{.}2}=10\textrm{.}4\times {{10}^{9}}=1\textrm{.}04\times {{10}^{10}} \\ & d=\sqrt{1\textrm{.}04}\times {{10}^{5}}=1\textrm{.}02\times {{10}^{5}}\ \text{m=102}\ \text{km} \\ \end{align}