The kinetic energy of a car is 1.02 × 105 J. If the car s speed increases by 28.2%, then what is the new kinetic energy of the car?

I assume $KE_0 = 1.02 \times 10^5 \, J$ ...
$KE_0 = \dfrac{1}{2}mv_0^2$
$KE_f = \dfrac{1}{2}m(1.282v_0)^2$
$\dfrac{KE_f}{KE_0} = \dfrac{\frac{1}{2}m(1.282v_0)^2}{\frac{1}{2}mv_0^2 }$
$\dfrac{KE_f}{KE_0} = \dfrac{(1.282v_0)^2}{v_0^2} = 1.282^2 \implies KE_f = KE_0(1.282^2) = 1.68 \times 10^5 \, J$