To find the value of \( b \) such that \( 6 \times 10^3 \) is 20 times larger than \( b \times 10^2 \), we can set up the equation:
\[ 6 \times 10^3 = 20 \times (b \times 10^2) \]
First, simplify the equation on the right:
\[ 6 \times 10^3 = 20b \times 10^2 \]
Next, we can divide both sides by \( 10^2 \) to eliminate it:
\[ \frac{6 \times 10^3}{10^2} = 20b \]
This simplifies to:
\[ 6 \times 10^{3-2} = 20b \]
Which further simplifies to:
\[ 6 \times 10^1 = 20b \]
Now \( 10^1 = 10 \), so we have:
\[ 6 \times 10 = 20b \]
Now, calculating the left side:
\[ 60 = 20b \]
Next, we can solve for \( b \) by dividing both sides by 20:
\[ b = \frac{60}{20} = 3 \]
Thus, the value of \( b \) that satisfies the equation is:
\[ \boxed{3} \]