Answer
$W_{f+a}=-1.21\times10^6J$
Work Step by Step
According to the work-energy theorem, $$\sum W_{nc}=E_f-E_0$$ $$\Big(\frac{1}{2}mv_f^2+mgh_f\Big)-\Big(\frac{1}{2}mv_0^2+mgh_0\Big)=\sum W_{nc}$$ $$\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)=\sum W_{nc}$$
The car's initial speed is $v_0=0$. We take the zero point to be at sea level, so $h_0=0$. From these, we can simplify the equation $$\frac{1}{2}mv_f^2+mgh_f=\sum W_{nc}$$
There are 3 non-conservative forces here: the propelling force $F$, friction and air resistance, so $\sum W_{nc}=W_F+W_{f+a}$, which means: $$W_{f+a}=\frac{1}{2}mv_f^2+mgh_f-W_F$$
We have $m=1.5\times10^3kg$, $h_f=2\times10^2m$, $v_f=27m/s$ and $W_F=4.7\times10^6J$. Therefore, $$W_{f+a}=-1.21\times10^6J$$
