A gas undergoes a cyclic process ABCA as depicted in the figure. If...

For a cyclic process, the change in internal energy is zero ($\Delta U = 0$).

According to the first law of thermodynamics, the net heat supplied is equal to the net work done:
$Q = W$

The net work done $W$ is the area enclosed by the cycle on the $P-V$ diagram.

From the given graph, the cycle forms a right-angled triangle with vertices at $A(100 \text{ kPa}, 500 \text{ cc})$, $B(200 \text{ kPa}, 500 \text{ cc})$, and $C(100 \text{ kPa}, 700 \text{ cc})$.

The base of the triangle is $\Delta P = 200 \text{ kPa} - 100 \text{ kPa} = 100 \times 10^3 \text{ Pa}$.

The height of the triangle is $\Delta V = 700 \text{ cc} - 500 \text{ cc} = 200 \times 10^{-6} \text{ m}^3$.

The work done is the area of this triangle:
$W = \dfrac{1}{2} \times \Delta P \times \Delta V$
$W = \dfrac{1}{2} \times (100 \times 10^3) \times (200 \times 10^{-6}) = 10 \text{ J}$

The net heat supplied is given as $Q = 2.4 \text{ cal}$.

Using the relation $W = JQ$, where $J$ is the mechanical equivalent of heat:
$10 = J \times 2.4$
$J = \dfrac{10}{2.4} = \dfrac{100}{24} \approx 4.17 \text{ J/cal}$

Answer: $4.17 \text{ J/cal}$