If the universe were shrunk to the size of the earth, then Earth be on this scale is :-

Magnification = size of earth / size of universe = 10^7 / 10^35 = 10^-18
Apprant size = 10^7 *10^18 = 10^-11 m (size of atom)..

**The Scale of the Universe**

To understand the size of the Earth on a scaled-down universe, let's first consider the actual size of the Earth and the universe.

The Earth has a diameter of roughly 12,742 km, while the observable universe is estimated to have a diameter of about 93 billion light-years, which is approximately 8.8x10^26 km. This vast difference in size makes it challenging to comprehend the scale we are dealing with.

**Shrinking the Universe**

Now, let's imagine shrinking the universe down to the size of the Earth. This means reducing the diameter of the universe from 8.8x10^26 km to 12,742 km. To determine the size of the Earth on this scale, we need to find the ratio between the original size of the universe and its shrunk size.

Ratio = Original Universe Size / Shrunk Universe Size

Ratio = 8.8x10^26 km / 12,742 km

Using scientific notation, we can express this ratio as:

Ratio = 8.8x10^26 km / 1.2742x10^4 km

Taking the ratio of the exponents, we subtract them:

Ratio = 8.8x10^(26 - 4)

Ratio = 8.8x10^22

**The Size of the Earth on a Shrunk Universe**

Now that we have the ratio, we can determine the size of the Earth on the scaled-down universe:

Earth's Size = Ratio x Earth's Diameter

Earth's Size = 8.8x10^22 x 12,742 km

Calculating this value, we get:

Earth's Size ≈ 1.12x10^27 km

Thus, on a shrunk universe, the Earth would have a size of approximately 1.12x10^27 km.

*Please note that the scale used in this explanation is a simplification for conceptual purposes. The actual scale and size of the universe are much more complex and vast, and this answer provides a simplified representation.*