Inversely proportional current controlled voltage source
simulate this circuit &ndash; Schematic created using CircuitLab

Instrumentation amplifier has very big input impedance. It measures the voltage on current sensing resitor (low impedance), then it is amplified G=1+2*Rf/Rg. Bias of 18V is applied, then you have to calculate Rf, RG so that amplified voltage is subrtracted from 18V to get 5V at xyz current.

The output impedance is low, but you can add a series resitor of your choice to match hi-impedance output.

\$V_{out}=18V-R_{sh}\cdot I\cdot (1+\dfrac{2\cdot R_f}{Rg})\$

\$R_{sh}=10; V_{sh}=10\cdot 8mA=80mV\$

\$\dfrac{dV_{out}}{dV_{in}}=\dfrac{18V-5V}{80mV}=162.5\$

\$(1+\dfrac{2\cdot R_f}{Rg})=162.5\$

\$R_f=2.2k\$

\$R_g=\dfrac{2\cdot R_f}{161.5} = 27.2\$

\$R=500\$