Answer
$5.23\;\rm m/s^2$
Work Step by Step
We know that, in general, the drive wheels are the rear wheels.
When the car starts moving the rear wheels push the road backward, so the road pushes back by the same amount of force forward [according to Newton's third law].
So, the net force exerted on the rear wheels on the concrete is due to the static friction force between the road and the drive wheels.
Thus,
$$\sum F_x=f_{s}=\mu_s F_n=m_{car}a_x$$
$$\mu_s F_n=m_{car}a_x\tag 1$$
and since two-thirds of the mass of the car is on the drive wheels, so the normal force exerted on them is given by
$$F_n=\dfrac{2}{3}m_{car}g$$
Plugging into (1) and solve for $a_x$;
$$ \dfrac{2}{3}\mu_sm_{car}g=m_{car}a_x$$
$$ \dfrac{2}{3}\mu_s g= a_x$$
Plugging the given
$$a_x= \dfrac{2}{3}\cdot 0.8\cdot 9.8=\color{red}{\bf 5.23}\;\rm m/s^2$$
