Boost Regulator issue at high load
Your inductor is awfully tiny for this application. For one thing, it needs to be rated to handle the input current, which is going to be more than 3&times; the output current.

Remember, the inductor current passes to the output only when the regulator's switch is off. Since you're going for a voltage multiplication of 3&times;, the switch duty cycle will be 2/3, and the diode will be conducting only 1/3 of the time. If you want to draw 200 mA at the output, the inductor is going to be carrying an average current of about 700 mA.

The switching freqeuncy of the regulator is fixed at 1.2 MHz. At 2/3 duty cycle, this means the switch is on for nominally 555 ns. With an input voltage of 4V and an inductor of 10 &micro;H, this means that the peak-to-peak ripple current is going to be

$$\Delta I = \frac{V}{L} \Delta t = \frac{4 V}{10 \mu H} 555 ns = 222 mA$$

This means that the peak current will be about 111 mA higher than the average current. You want the saturation current of the inductor to be a comfortable margin above the peak value.

When your inductor saturates, you'll see the exact symptoms you're getting: poor regulation and poor efficiency.

Also, with a peak current of around 800 mA, even the 0.34&Omega; of series resistance is going to cut into your efficiency with a drop of almost 300 mV, which is nearly 10% of your input voltage.