Since this is a circuit theory question, it's possible the question author wants us to assume the inductors are ideal, that is they have zero series resistance.
If the inductors initially have 0 current through them1 as the switch is closed, and the voltage across them is $v(t)$, then the current through the first is
$$i_1(t) = \int_0^t \frac{v(t)}{L_1}\rm{d}t$$
And across the second one
$$i_2(t) = \int_0^t \frac{v(t)}{L_2}\rm{d}t$$
Because the integral is a linear operator, we can pull out the constant term and find
$$L_1 i_1(t) = L_2 i_2(t)$$
for any choice of $t$, including in the limit as $t\to\infty$.
Therefore in the steady state we find2
$$I_1 = \frac{L_2}{L_1}I_2$$
From this you can find the current through the individual inductors in the situation given. (Hint: the answer will be very similar to if they were two resistors with values $R_1$ and $R_2$)
Note 1: This is actually not a great assumption for the circuit as given, because when the switch is open there is no current flow through the resistor, and therefore no mechanism for the inductors to lose energy and decay to a 0-current state when the switch is open.
Note 2: In a real world circuit, however, the steady state behavior would be dominated by the equivalent series resistance of the two inductors, and their inductance would not come in to play. See rob's answer for more details.