Get practial resistances for the Current Divider
Since there's the same voltage across both resistors, then if there's \$ 40\mu A\$ through Rl and \$ 60\mu A\$ through Rs, Rl must be equal to 1.5 times Rs.

Your job, then, is to pick from the list, two resistors with a resistance ratio of 1.5 which when paralleled present an equivalent resistance of 60k\$\Omega\$. Or 80k\$\Omega\$. Or anything in between...

Looking for a 60k\$\Omega\$ solution,

If we set \$ Rs = 1 \text { and } R_l = 1.5\$, we can write:

$$k= \frac{R_s\times R_l}{R_s+R_l}= \frac{1.5}{2.5}=0.6 $$

and then:  $$Rs =\frac{Rt}{k} = \frac{60k\Omega}{0.6} = 100k\Omega $$

and, since the resistance of Rl is 1.5 times that of Rs:

$$ R_L = 1.5 \times R_s = 150k\Omega$$ 

Consulting the list, we find 10 and 15, so we lucked out with a perfect match!