Current probe settings and measurement
If you assumed that the current was really peaking at 7.8 amps AND, that the inductor is 68 uH (as per your comment) AND, that it is used in a boost converter as the energy storage inductor then, you could calculate what the current was (providing you know the power supply voltage).

But, assuming that the current is 7.8 amps and that it rises from zero amps in 50 us to 7.8 amps, you can back-calculate the supply. Firstly the rise time of the current: -



Based on your scope setting, I estimate the rise time to be about 50 us. This means that \$\frac{di}{dt}\$ = 7.8/(50E-6) = 156,000 amps per second.

Multiply this by inductance and you get a measure of what the input voltage was during the measurement you made.

So, \$V= L\frac{di}{dt}\$ = 10.6 volts.

If your input supply was around this value then, the current your probe measured is probably correct and you should rethink the inductor you have used because it will get hot and ruin the efficiency of your boost conversion in some cases.