Answer
${\bf 2.8}\;\rm mA$
Work Step by Step
Let's assume that the muscle and fat are ohmic resistors.
We know, according to Ohm's law, that
$$\Delta V=IR$$
where $R=\rho L/A$ and $A=\pi r^2=\pi D^2/4$ where $D$ is the diameter.
$$\Delta V=I\dfrac{4\rho_{\rm } L}{\pi D^2}$$
Thus,
Solving for $I$;
$$I=\dfrac{\pi D^2\Delta V}{4\rho L}\tag 1$$
We know that the upper leg consists of 82% muscle and 18% fat, so
$$\rho=0.82\rho_{\rm muscle }+0.18\rho_{\rm fat}$$
Plug into (1),
$$I=\dfrac{\pi D^2\Delta V}{4L (0.82\rho_{\rm muscle }+0.18\rho_{\rm fat})} $$
Plug the known;
$$I=\dfrac{\pi (0.12)^2(1.5)}{4(0.40) (0.82[13]+0.18[25])} $$
$$I=\color{red}{\bf 2.8}\;\rm mA$$
