DC Boost Converter stepping up voltage very slowly, if any
Let's calculate what happens when we turn your transistor on for 8.3 ms.

We know that inductors have the constituent equation

$$\frac{dI}{dt} = \frac{V}{L}$$

So when we apply 5 V to one side of L1 and ground the other side, we get

$$\frac{dI}{dt} = \frac{5\ {\rm V}}{16\ {\rm\mu H}}= 312,500\ \frac{\rm A}{\rm s}$$

And after 8.3 ms, the current should ramp to

$$(312,500\ \frac{\rm A}{\rm s})(8.3\ {\rm ms})=2,593\ {\rm A}$$

This is much more than your 5 V supply is likely to be able to supply, and much more than your inductor is able to carry without vaporizing.

You need to design your circuit for much, much smaller current ripple through the inductor.

You could look for a 16 mH inductor able to support 3 A current (which will be big and expensive). You will likely need to increase your load capacitance by a few orders of magnitude as well to achieve acceptable voltage ripple.

Or you could use a much higher switching frequency. Typical switching frequencies for this kind of circuit are 50 kHz to 5 MHz, so at least 500 times higher than your 0.1 kHz.

You might need to use a built-for-purpose switching controller instead of a general-purpose microcontroller to achieve the switching frequency you need.