An upper wire hangs from a ceiling and supports a mass m_2 = 36 kg...

Let $m$ be the mass of the load added to the hanger.

The tension in the lower wire is given by the total weight it supports:
$T_1 = (m_1 + m)g = (10 + m) \times 10$

The tension in the upper wire is given by the total weight it supports:
$T_2 = (m_2 + m_1 + m)g = (36 + 10 + m) \times 10 = (46 + m) \times 10$

The maximum tension the lower wire can withstand before breaking is:
T_{1, \text\max} = \text{Breaking stress} \times A_1
T_{1, \text\max} = (8 \times 10^8) \times (0.003 \times 10^{-4}) = 240 \text{ N}

The maximum tension the upper wire can withstand before breaking is:
T_{2, \text\max} = \text{Breaking stress} \times A_2
T_{2, \text\max} = (8 \times 10^8) \times (0.006 \times 10^{-4}) = 480 \text{ N}

For the lower wire not to break:
$(10 + m) \times 10 \le 240$
$10 + m \le 24 \Rightarrow m \le 14 \text{ kg}$

For the upper wire not to break:
$(46 + m) \times 10 \le 480$
$46 + m \le 48 \Rightarrow m \le 2 \text{ kg}$

The maximum load that can be added without breaking either wire is the smaller of the two values, which is $2 \text{ kg}$.

If the load is further increased beyond $2 \text{ kg}$, the tension in the upper wire will exceed its maximum capacity first. Therefore, the upper wire will break first.