According to the universal law of gravitation, gravitational force exerted on an object of mass $m$ is given by

$F=\frac{G M m}{r^{2}}$

Where,

Mass of Earth, $M = 6 \times 10^{24}\, kg$

Mass of object, $m = 1 \,kg$

Universal gravitational constant, $G = 6.7 \times 10^{-11}\, Nm^2 \,kg^{-2}$

Since the object is on the surface of the Earth,

$r =$ radius of the Earth $(R)$

$r = R = 6.4 \times 10^6\, m$

Therefore, the gravitational force

$F=\frac{G M m}{r^{2}}=\frac{6.7 \times 10^{-11} \times 6 \times 10^{24} \times 1}{\left(6.4 \times 10^{6}\right)^{2}}=9.8\, N$