We have drawn the figure according to the information given in the question,

Given information:
- Smita rode her scooty Northwards initially.
- Then she turned left.
- Then she rode to her left for 4 km.
- She found herself exactly 2 km West of her starting point.

To find:
The distance Smita rode Northwards initially.

Solution:
Let's analyze the given information step by step.

Step 1: Smita rode her scooty Northwards initially.
This means that her initial direction of travel was towards the North.

Step 2: Then she turned left.
When she turned left, her direction of travel changed by 90 degrees anticlockwise. So, now she is heading towards the West.

Step 3: Then she rode to her left for 4 km.
Since she turned left, she rode 4 km towards the South.

Step 4: She found herself exactly 2 km West of her starting point.
This means that after traveling 4 km South, she ended up 2 km West of her starting point.

Visual representation:
Let's represent Smita's initial position as point A. When she traveled Northwards, she reached point B. Then, after turning left and traveling 4 km to her left (Southwards), she ended up at point C, which is 2 km West of her starting point.

C
|
|
|_________ 2 km
| |
| |
| |
| |
A --> B

Calculating the distance Smita rode Northwards initially:
To find the distance Smita rode Northwards initially, we need to calculate the vertical distance between points A and B.

Given that she traveled 4 km Southwards and ended up 2 km West of her starting point, we can use Pythagoras' theorem to find the vertical distance.

Using Pythagoras' theorem:
AC^2 = AB^2 + BC^2

Substituting the known values:
AC^2 = (distance Smita rode Northwards initially)^2 + 4^2

Since she ended up 2 km West of her starting point, BC = 2 km.

AC^2 = (distance Smita rode Northwards initially)^2 + 4^2
AC^2 = (distance Smita rode Northwards initially)^2 + 2^2

Given that AC = distance Smita rode Northwards initially.

(distance Smita rode Northwards initially)^2 = AC^2 - 2^2
(distance Smita rode Northwards initially)^2 = AC^2 - 4
(distance Smita rode Northwards initially)^2 = (AC^2 - 2^2)
(distance Smita rode Northwards initially)^2 = (AC^2 - 4)

Since AC^2 is greater than 4, the distance Smita rode Northwards initially must be greater than 2.

Hence, the correct answer is option B) 4 km.