To transition a system from state A to state B along the path ACB...

The work done along the path $ACB$ is the sum of the work done during the isobaric process $AC$ and the isochoric process $CB$.

$W_{ACB} = W_{AC} + W_{CB}$

$W_{AC} = p_A (V_C - V_A) = 50 \times 10^3 \times (400 - 200) \times 10^{-6} = 10 \text{ J}$

$W_{CB} = 0$ (since volume is constant)

$W_{ACB} = 10 \text{ J}$

The work done along the path $ADB$ is the sum of the work done during the isochoric process $AD$ and the isobaric process $DB$.

$W_{ADB} = W_{AD} + W_{DB}$

$W_{AD} = 0$ (since volume is constant)

$W_{DB} = p_D (V_B - V_D) = 155 \times 10^3 \times (400 - 200) \times 10^{-6} = 31 \text{ J}$

$W_{ADB} = 31 \text{ J}$

According to the first law of thermodynamics, the change in internal energy $\Delta U$ is path-independent.

$\Delta U = Q_{ACB} - W_{ACB} = Q_{ADB} - W_{ADB}$

$Q_{ADB} = Q_{ACB} + W_{ADB} - W_{ACB}$

Substituting the given values and the calculated work done:

$Q_{ADB} = 50 \text{ cal} + (31 \text{ J} - 10 \text{ J}) = 50 \text{ cal} + 21 \text{ J}$

Using the mechanical equivalent of heat $1 \text{ cal} = 4.2 \text{ J}$:

$21 \text{ J} = \dfrac{21}{4.2} \text{ cal} = 5 \text{ cal}$

$Q_{ADB} = 50 \text{ cal} + 5 \text{ cal} = 55 \text{ cal}$