From Mechanics

The Law of gravitation applied to a particle on Mars gives,

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

where \displaystyle F is the force on a particle on the surface of Mars, \displaystyle {{m}_{1}} is the mass of Mars, \displaystyle {{m}_{2}} is the mass of the particle and \displaystyle d is the radius of Mars.

As \displaystyle F={{m}_{2}}a where \displaystyle a is the acceleration of the particle on Mars,

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

The radius of Mars must be expressed in SI units.

\displaystyle d=3\textrm{.}4\times {{10}^{6}}\ \text{m}

and 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} & a=\frac{6\textrm{.}67\times {{10}^{-11}}\text{ k}{{\text{g}}^{\text{-1}}}{{\text{m}}^{\text{3}}}{{\text{s}}^{\text{-2}}}\times \text{ 6}\textrm{.}\text{42}\times \text{1}{{0}^{\text{23}}}\text{kg}}{{{\left( \text{3}\textrm{.}\text{4}\times \text{1}{{0}^{\text{6}}}\text{m} \right)}^{2}}} \\ & =\frac{6\textrm{.}67\times 6\textrm{.}42}{{{3\textrm{.}4}^{2}}}\ \text{m}{{\text{s}}^{-2}}=3\textrm{.}7\ \text{m}{{\text{s}}^{-2}} \\ \end{align}