An Atwood machine has two blocks of masses 2.0 kg and 3.0 kg ....

The acceleration of the blocks in the Atwood machine is given by:

$a = \dfrac{m_2 - m_1}{m_1 + m_2} g$

Substituting the given masses $m_1 = 2.0 \text{ kg}$ and $m_2 = 3.0 \text{ kg}$:

$a = \dfrac{3.0 - 2.0}{3.0 + 2.0} g = \dfrac{g}{5}$

The displacement of the blocks during the $4^{\text{th}}$ second is:

$s_4 = u + \dfrac{a}{2}(2n - 1)$

$s_4 = 0 + \dfrac{g/5}{2}(2(4) - 1) = \dfrac{7g}{10}$

The net work done by gravity on the system is the sum of the work done on each block. Since the heavier block moves downwards and the lighter block moves upwards:

$W = m_2 g s_4 - m_1 g s_4 = (m_2 - m_1) g s_4$

$W = (3.0 - 2.0) g \left(\dfrac{7g}{10}\right) = \dfrac{7g^2}{10}$

Taking standard gravity $g = 9.8 \text{ m/s}^2$:

$W = \dfrac{7 \times (9.8)^2}{10} = \dfrac{7 \times 96.04}{10} = 67.228 \text{ J}$

Rounding to the nearest integer, we get $W \approx 67 \text{ J}$.