I assume you mean that the battery is 16 kWh.
If you're going to make a boost converter, buy a dedicated boost converter IC, rather than using a 555. You could put all the panels in series, which would make the voltage conversion easier.
Why do you need at least 90% efficiency? You will struggle to get that high and the input power is free. You could just have another panel or two. Also, you will get a good efficiency improvement with an MPPT circuit (maximum power point tracking) because the power from the solar cells will vary with sunlight intensity and temperature. There are circuits that will maximise the charging rate, but they are complicated.
There are many issues with charging an electric car. I guess you must be charging a plug-in hybrid, as 16 kWh is too small for a purely electric vehicle.
All purely electric vehicles (Nissan Leaf, Teslas, Jaguar iPace etc) and, I assume, the plug-in hybrids, need 12 V to be applied to the high voltage battery to get the contactors to turn on before charging can take place. You will need to get the car's systems to turn the contactors on, or bypass them completely, for charging to happen. Bypassing the systems may lead to the car's electronics misunderstanding the state of charge of the battery. A 330 V battery has many cells in series and you would want to make sure that the cell balancing is running when you are charging.
The 400 V systems on cars are completely isolated from the 12 V systems. The negative of the 12 V system is connected to the car body, while the 400 V system is entirely isolated, in the same way that your phone or laptop charge is isolated from the mains. You can't go connecting 12 V panels to the 400 V systems as there is the possibility of the solar panels being at 400 V compared to the car's body. So you will need an isolated converter to remove that danger.
I'm trying to work out what you are attempting to achieve with this. Gasoline has an energy density of around 132 MJ/usgallon. If the engine charges the battery with an efficiency of 20%, that comes to 26.5 MJ of electricity per US gallon, or about 7.5 kWh. So you would use around 2 gallons of gasoline to charge the battery, but probably less if you were driving, as regenerative braking will charge it as well. So that is a $6 saving in 2 days of charging, or more like 4 days as the sun only shines half the time. It is hard to see how you will get a useful payback time.