The figure is not visible in the prompt, so the numerical values of the two self-inductances, the two resistances, the mutual inductance, the battery voltages, and the relative polarity/dot convention are missing. Without those data, the total work done by the batteries against induced emf cannot be uniquely calculated from first principles.

However, the required physics result is:

1. Energy stored in two mutually coupled inductors

If the final steady currents are $I_1$ and $I_2$, then the total magnetic energy stored is $U = \frac{1}{2}L_1 I_1^2 + \frac{1}{2}L_2 I_2^2 \pm M I_1 I_2,$ where the sign depends on whether the mutual coupling is aiding or opposing according to the dot convention.

2. Why this equals the work done against induced emf

In an LR circuit, the battery does work partly against resistor dissipation and partly against the back emf of the inductors. The total work done against induced emf from $t=0$ to steady state equals the final magnetic energy stored in the coupled inductors. Hence, $W_{\text{against induced emf}} = \frac{1}{2}L_1 I_1^2 + \frac{1}{2}L_2 I_2^2 \pm M I_1 I_2.$

3. Steady-state currents

At steady state, inductors behave as short circuits, so $I_1 = \frac{V_1}{R_1}, \qquad I_2 = \frac{V_2}{R_2}$ for the two loops, assuming simple series LR branches.

Thus the numerical answer should be $W = \frac{1}{2}L_1\left(\frac{V_1}{R_1}\right)^2 + \frac{1}{2}L_2\left(\frac{V_2}{R_2}\right)^2 \pm M\left(\frac{V_1}{R_1}\right)\left(\frac{V_2}{R_2}\right).$

To convert into mJ, multiply joules by $10^3$.

4. Comparison with stored answer

Since the figure values are missing, I cannot independently derive the numeric value $55\,\text{mJ}$ from the given prompt alone. The stored answer may well be correct for the unseen circuit values, but the derivation cannot be completed uniquely without them.

So, based on the information available:

• the method is certain,
• the numeric verification is not possible from the text alone.