Vccs op-amp with transistor
Vcc need to be chosen based on a few things.

You need to examine the spec-sheet of the particular op-amp.

First: The device itself will have a minimum voltage at which it will operate so your VCC needs to be at least that large.

Second: the op-amp needs to output a voltage at least as large as the maximum control voltage plus the base emitter diode drop in that transistor. Op-Amps have a voltage swing that does not go all the way up to Vcc. Look at the spec-sheet and see how much greater Vcc needs to be over the voltage you just calculated.

Whichever of those voltages is greater is your minimum Vcc. Chose a suitable supply greater than that value.

Vcc- could be ground in this circuit since you do not need negative voltages. However, not all op-amps will work rail to rail and not all op-amps will drive all the way to the bottom rail. Again, you need to check the data sheet to see what you need on this supply so your output can reach zero.

In this circuit, since the current through the sense resistor is \$Ic + Ib\$  then collector current is actually governed by the equation

\$Ic = (V_{Control}/R_{Sense}) * H_{fe}/(H_{fe}+1) \$

As such, since \$H_{fe}\$ is quite variable, it is not that great a circuit.

ADDITION: 

To calculate the \$V\$ value, you need to know the load resistance. This is where your example schematic is incomplete. At this point you have to assume the implied components shown below exist.







simulate this circuit &ndash; Schematic created using CircuitLab

\$V_{min} = V_{ControlMax} + I_{cMax} * R_{Load} + V_{CESat}\$

However it can not be grater than the max \$V_{CE}\$ of the transistor nor high enough to cause the transistor to overheat from \$P = V_{CE} * I_{cMax}\$

\$V_C\$ will of course be \$V - (R_{LOAD} * I_C)\$