What I mean is, if you look at the voltages between op-amp output, -in, and the top of the shunt resistor, you see: 30p, 10n -- a capacitive voltage divider, with rather low gain. Which means the op-amp won't be seeing much of its own output, and won't have good phase margin. Eliminating the 10n is equivalent to setting it to ~0p, so the op-amp -in sees 30p from output and the phase margin is good.
Tim