Answer
$40\;\rm nA$
Work Step by Step
Let's assume that the axon can be represented as a long ideal current-carrying wire.
So the magnetic field created by it is given by
$$B=\dfrac{\mu_0I}{2\pi d}$$
We need the peak current, so
$$I=\dfrac{2\pi Bd}{\mu_0}$$
Plug the known;
$$I=\dfrac{2\pi (8\times 10^{-12})(1\times 10^{-3})}{(4\pi \times 10^{-7})}$$
$$I=\color{red}{\bf 4\times 10^{-8}}\;\rm A$$