I am design a transistor switch circuit.
OK, that then means that we are going to operate the NPN in saturation mode.
Why not in Active mode?
Because we want a switch, in active mode the transistor doesn't act as a switch it acts more like a variable current source.
In active mode we use the transistor's high current gain \$\beta = \frac {I_c}{I_b}\$. That then means that \$I_b\$ controls \$I_c\$. But we don't want "control", we want a switch so: on / off that's it.
That's where saturation mode comes in. In saturation mode we simply make \$I_b\$ so large that the \$I_c = \beta * I_b\$ becomes much larger than the actual \$I_c\$ that is flowing.
What then determines \$I_c\$? In saturation mode I mean.
The \$I_c\$ in saturation mode is then determined by the load. Your load wants 30 mA to flow so we need to make sure that we apply enough \$I_b\$ to the transistor so that the 30 mA can easily flow.
In active mode to make 30 mA flow when \$\beta\$ = 100 we'd need \$I_b\$ = 0.3 mA. But as I mention above, to make sure we're in saturation mode we need to apply a much higher \$I_b\$ than that!
How much more, well that is a choice, in the datasheet of the PDTC123J, figure 7 they chose \$\frac {I_c}{I_b}\$ = 20. I emphasize chose because it is a choice, they could also have chosen \$\frac {I_c}{I_b}\$ = 30. As long as the value is significantly smaller than \$\beta\$ (which is a factor 100 or more for this transistor) the transistor will be in saturation mode.
So for your load's 30 mA and chosing \$\frac {I_c}{I_b}\$ = 20 that would mean you need to make
\$I_b = \frac{I_c}{20}\$ = 30 mA / 20 = 1.5 mA
If you would chose \$\frac {I_c}{I_b}\$ = 30 then you would need \$I_b\$ = 1 mA and the transistor would be a little less deep into saturation. That would mean there would be slightly more voltage drop across the transistor when it is switched on.
Edit
You asked about Figure 6 which shows the DC current gain as a function of collector current
This plot is relevant as it shows what the transistor's \$\beta\$ is for a certain collector current. For your application we know that the collector current is 30 mA. From Figure 6 we can then see that \$\beta\$ (the plot shows \$h_{fe}\$ which is just a different name for \$\beta\$) is always more than a about 150.
As described above, for saturation mode we need \$\frac {I_c}{I_b} << \beta\$ and that condition is met in my examples above.
Note that you need a very small value of \$I_c\$ for \$\beta\$ to become much smaller. Experienced engineers know this so ignore the plot and just use the value of \$\beta\$ from the tables.
