Three masses m_1 = 4 kg, m_2 = 4 kg and m_3 = 6 kg are suspended...

Let the acceleration of the system be $a$. The total mass on the right side is $m_2 + m_3 = 4 + 6 = 10$ kg, and the mass on the left side is $m_1 = 4$ kg. Since $10$ kg $> 4$ kg, the right side accelerates downwards and the left side accelerates upwards.

The common acceleration $a$ of the system is given by:
$a = \dfrac{(m_2 + m_3) - m_1}{m_1 + m_2 + m_3} g$
$a = \dfrac{10 - 4}{4 + 4 + 6} g = \dfrac{6}{14} g = \dfrac{3}{7} g$

For mass $m_1$, the equation of motion is:
$T_1 - m_1 g = m_1 a$
$T_1 = m_1(g + a) = 4 \left(g + \dfrac{3}{7}g\right) = 4 \left(\dfrac{10}{7}g\right) = \dfrac{40}{7}g$

For mass $m_3$, the equation of motion is:
$m_3 g - T_2 = m_3 a$
$T_2 = m_3(g - a) = 6 \left(g - \dfrac{3}{7}g\right) = 6 \left(\dfrac{4}{7}g\right) = \dfrac{24}{7}g$

The ratio of the tensions is:
$\dfrac{T_1}{T_2} = \dfrac{\dfrac{40}{7}g}{\dfrac{24}{7}g} = \dfrac{40}{24} = \dfrac{5}{3}$

Answer: $5/3$