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Let's first extract the given values for \( A \) and \( B \) and calculate their ratio \( \frac{B}{A} \).

The expressions provided are:

\[
A = 0.2 \cdot 10^{-6} + 3 \cdot 10^{-7} - 0.01 \cdot 10^{-5}
\]
\[
B = 6 \cdot 10^2 - 0.02 \cdot 10^4
\]

### Step 1: Simplify the values for \( A \) and \( B \)

#### For \( A \):
\[
A = 0.2 \cdot 10^{-6} + 3 \cdot 10^{-7} - 0.01 \cdot 10^{-5}
\]
Convert all terms to the same exponent base:

\[
0.01 \cdot 10^{-5} = 0.1 \cdot 10^{-6}
\]

Thus, we have:
\[
A = 0.2 \cdot 10^{-6} + 3 \cdot 10^{-7} - 0.1 \cdot 10^{-6}
\]
Combine the terms with \( 10^{-6} \):

\[
A = (0.2 - 0.1) \cdot 10^{-6} + 3 \cdot 10^{-7} = 0.1 \cdot 10^{-6} + 3 \cdot 10^{-7}
\]
\[
A = 1 \cdot 10^{-7} + 3 \cdot 10^{-7} = 4 \cdot 10^{-7}
\]

#### For \( B \):
\[
B = 6 \cdot 10^2 - 0.02 \cdot 10^4
\]
Convert the second term to match powers of 10:
\[
0.02 \cdot 10^4 = 2 \cdot 10^2
\]
Thus, we have:
\[
B = 6 \cdot 10^2 - 2 \cdot 10^2 = (6 - 2) \cdot 10^2 = 4 \cdot 10^2
\]

### Step 2: Calculate the ratio \( \frac{B}{A} \)

\[
\frac{B}{A} = \frac{4 \cdot 10^2}{4 \cdot 10^{-7}} = 10^{2 + 7} = 10^9
\]

Thus, the correct answer is \( \boxed{10^9} \) (option D).

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Would you like any further clarification or details?

Here are 5 related questions you might find useful:
1. How can you simplify expressions with scientific notation?
2. How do you convert between different powers of 10 in calculations?
3. What are common mistakes to avoid when dealing with very small or large numbers in scientific notation?
4. How does adding and subtracting values affect scientific notation terms?
5. How can scientific notation help in comparing very large and very small numbers?

**Tip:** Always simplify terms to the same power of 10 before performing addition or subtraction in scientific notation.

Given A = 0.2 * 10^-6 + 3 * 10^-7 - 0.01 * 10^-5 and B = 6 * 10^2 - 0.02 * 10^4, what is the ratio B/A?

Learn how to solve a scientific notation problem involving the ratio of B/A, where A and B are given in powers of 10. Step-by-step simplification provided, suitable for students in Grades 9-12.

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