It is not hard calculating the values of the biasing resistors. But before I do it, I like to point out that if there is a mechanical switch, the transistor is not needed. The anode of the external LED will be connected to VCC1 and R3 is connected to the switch J1 instead of ground.
Now let us be back to transistor Q1.
We start here from knowing its Ic (collector current) when it is on:
Ic = (VCC1 - Vce_sat - V_led1 - V_ir) / R3 , [equation_1]
VCC1 = 5V, [given_1]
Vce_sat = 0.2 V , estimated value of the saturated Vce, [estimated_1]
V_led1 = 2 V , typical value of red LED, [estimated_2]
V_ir = 1.2 V , typical value of IR LED (the opto-coupler input), [estimated_3]
R3 - 100 Ohm, [given_2]
Ic = 16 mA , [calculated_1]
This current is relatively high for the 1mA load at the output of the opto-coupler.
If you like to reduce it, say to 2 mA, you can also use [equation_1] to calculate R3 and choose the closest standard value (lower of the result).
In this case (saturated Q1), the ratio Ic/Ib could be estimated about 20.
G_i = Ic / Ib = 20 , [estimated_4]
Therefore:
Ib = Ic / G_i = 16 / 20 = 0.8 mA , [calculated_2]
We also assume the current in R1 from Ib*0.1 to 1*Ib. A higher current (I_r1) lets R1 be smaller and this gives faster turn off and better immunity to noise. Here even I_r1 = Ib/10 is fine but let us assume:
I_r1 / Ib = 0.5 , [estimated_5]
From [calculated_1]:
I_r1 = 0.5 * 0.8 = 0.4 mA , [calculated_3]
But:
R1 = Vbe_q1 / I_r1 , [equation_2]
where:
Vbe_q1 = 0.7 , the diode forward volatge , [estimated_5]
R1 = 0.7 / 0.4 = 1.75 Kohm , [calculated_4]
R1 = 1.8 Kohm , [result_1]
The actual I_r1 can be found with [equation_2]:
I_r1 = 0.7 / 1.8 = 0.389 mA , [calculated_5]
Now:
I_r2 = Ib + I_r1 , [equation_3]
I_r2 = 0.8 + 0.389 = 1.189 mA , [calculated_6]
And:
R2 = ( VCC1 - Vbe_q1 ) / I_r2 , [equation_4]
R2 = ( 5 - 0/7 ) / 1.189 = 3.62 Kohm , [calculated_7]
R2 = 3.3 Kohm , [result_2]
Hope this helps