Calculationg base-resistor between microcontroller and transistor
You are missing one easy thing: voltage across the resistor won't be 5V.

Transistor side you have a \$V_{BE}=V_\gamma\approx0.7V\$, so the base of the resistor is about 0.7V lifted from ground.

Microcontroller side, when the output is high you don't have full \$V_{CC}\$ but a voltage that is somewhat lower. For a cmos chip that can be quite low, some 100mV, but to make the calculations easier let's say you have 0.3V from \$V_{CC}\$, meaning that the arduino output voltage is only 4.7V.

Let's do the math again:

$$R=\frac{(5-0.3-0.7)\text{V}}{10\text{mA}}=400\Omega$$

Since higher base current is better in this case, just stick with the lowest nearest standard value, i.e. \$390\Omega\$.

Note: that 0.3V that I wildly guessed is actually written down in the microcontroller datasheet and it's called \$\mathbf{V_{OH}}\$, as in Voltage Output High. You also can find the \$\mathbf{V_{OL}}\$ and... Yes, you guessed it, that's Voltage Output Low.
The first one is the minimum output voltage when the pin is set high, if you are within current specs of course, while the second value is the maximum output voltage when the pin is set low.