Linear Transformer and Ideal Transformer: electric circuit question
Then, how can I find the voltage across the 30mH winding


For coupled inductors we have the following:

$$M = k\sqrt{L_1 L_2}$$

$$V_1 = L_1 \dot I_1 + M \dot I_2$$

$$V_2 = M \dot I_1 + L_2 \dot I_2$$

Now, if \$I_2\$ is constant, we have:

$$V_1 = L_1 \dot I_1$$

$$V_2 = M \dot I_1$$

Thus:

$$V_2 = M \dfrac{V_1}{L_1} =V_1 \cdot k\sqrt{\frac{L_2}{L_1}} $$



The remainder of the answer was reasoned out and added below by the OP...

Now, to find the voltage across the 400 winding:

$$\frac{V_a}{V_b}=\frac{1600}{400}=4$$
where 
\$V_a \$ is the voltage across the 1600 winding and \$V_b\$ is the voltage across the 400 winding.
We also can see that, 
$$V_a+V_b=V_1$$
$$ \implies V_b=\frac{1}{4}(V_1-V_b)$$
$$V_b=\frac{1}{5}V_1$$
But to have the current in \$Z_x\$ as zero, we need to have \$V_b=V_2\$. So,
$$V_1 \cdot k\sqrt{\frac{L_2}{L_1}}=\frac{1}{5}V_1$$
If we solve, the above equation, we get
$$k=\frac{2}{5}=0.4$$