$m=1800 \mathrm{~kg}$.
Let $R_1, R_2=$ Reaction forces due to ground on each front wheel and each back wheel.
$$
\therefore \quad R_1+R_2=\mathrm{mg}=1800 \times 9.8
$$
Distance between the front and back axles $=1.8 \mathrm{~m}$
Distance of centre of gravity from the front axle $=1.05 \mathrm{~m}$
For rotational equilibrium of the car about the centre of gravity,
$$
\begin{aligned}
&R_1 \times 1.05=R_2(1.8-1.05)=R_2 \times 0.75 \\
&\frac{R_1}{R_2}=\frac{0.75}{1.05}=\frac{5}{7}
\end{aligned}
$$
From (1) and (2), $\frac{5}{7} R_2+R_2=1800 \times 9.8$
$$
\begin{gathered}
\Rightarrow R_2=\frac{1800 \times 9.8 \times 7}{12}=10290 \mathrm{~N} . \\
R_1=\frac{5}{7} R_2=\frac{5}{7} \times 10290=7350 \mathrm{~N}
\end{gathered}
$$