**Answer:**

To find out the load required to stretch the brass wire by 1cm, we need to consider the properties of the wire and utilize Hooke's law.

**Hooke's Law:**
Hooke's law states that the force required to extend or compress a spring-like object is directly proportional to the extension or compression of the object. Mathematically, it can be expressed as:

F = k * x

Where:
F is the force applied,
k is the spring constant (also known as the modulus of elasticity),
x is the extension or compression of the object.

**Given Data:**
Length of the brass wire (L) = 8m
Diameter of the wire (d) = 0.4mm
Extension of the wire (x) = 1cm = 0.01m
Young's modulus of brass (Y) = 90 GPa = 90 * 10^9 Pa

First, we need to calculate the cross-sectional area of the wire using its diameter:

**Calculating Cross-sectional Area:**
The diameter of the wire is given as 0.4mm, which can be converted to meters by dividing it by 1000.
Radius (r) = d/2 = (0.4/1000)/2 = 0.0002m

The cross-sectional area (A) of the wire can be calculated using the formula:
A = π * r^2
A = π * (0.0002)^2 = 0.0000001257 m^2

**Calculating Spring Constant:**
The spring constant (k) can be calculated using the Young's modulus (Y) and the cross-sectional area (A) of the wire:
k = Y * A / L

Substituting the given values:
k = (90 * 10^9 Pa) * (0.0000001257 m^2) / 8m
k = 0.00141375 N/m

**Calculating Load:**
Now, we can use Hooke's law to calculate the load (F) required to stretch the wire by 1cm:
F = k * x
F = (0.00141375 N/m) * 0.01m
F = 0.0000141375 N

Therefore, the load required to stretch the brass wire by 1cm is approximately 0.0000141375 N.

Note: The above calculations assume that the wire is perfectly elastic and obeys Hooke's law within the elastic limit.