Sounds like a competitor for the Keysight CX3300 family, arguably the leader in this space. (If you have to ask how much it costs, you can’t afford it.)
But the real question is how much bandwidth do you need? If you are measuring current directly into an IC without any bypass capacitance, you can make use of this large measurement bandwidth.
If you are measuring the current to a target circuit board, not so much. You have the impedance from your power supply, cables/wires, and any connectors. If you do a very good job with very short wires and super low-impedance supply, this could be 2.5mΩ in each direction, + and -. With even just 10 µF bypass capacitance on your board, your target board bandwidth is:
1 / (2 * π * 5e-3 Ω * 10e-6 µF) = 3 MHz.
Most typical lab setups with banana jacks and 3’ of wires will be closer to 50 mΩ. Banana jacks have 1 mΩ connector resistance. 18 AWG is 6.3 mΩ/ft.
4 connectors + 6 ' of 18 AWG = 4 * 1 mΩ + 6.3 mΩ/ft * 6 ft = 42 mΩ
With this setup, you only get 379 kHz of system bandwidth. Coincidentally, this is how I determined the design objective for Joulescope’s target bandwith. Joulescope uses a 10 mΩ shunt in its highest current range setting. Other impedances (Joulescope’s MOSFET, PCB traces, the internal front-panel connector and the external connectors) add another ~15 mΩ. So, with 2 sets of 1’ banana wires (Joulescope sits between your power supply and target board), the system input resistance is:
4 connectors * 1 mΩ + 4' * 6.3 mΩ + 25 mΩ = 54.2 mΩ
This setup with 10 µF target bypass capacitance gives 294 kHz which is slightly higher than Joulescope’s 250 kHz bandwidth specification, but that spec is a conservative number.
Note that I am intentionally leaving out the supply impedance. A small LiPo battery can easily add 100s of mΩ.
So, if you want to measure current to a target board, save your $$$ and buy a Joulescope . Joulescope’s sensor inputs are electrically isolated, up autorange in 1 µs (no target brown-out), and measure 32-bits of effective range.