Super Calculating

We know: $101\times 99 = 9999$

$1001\times 999=999999$

$10001 \times 9999 = 99999999$

So we can say that: multiplying $100....001$ (n zeros) by $99...9$ (n+1 nines) equals $99...99$ (2n+2 nines)

By using that fact, we can convert A into a decimal number with 300 decimal digits:

First, we should eliminate the dot by multiplying both the numerator and denominator by $10^{100}$ . Therefore, A= $\frac{10^{100}}{100...01}$ (99 zeros)

Second, we multiply both the numerator and denominator by $(10^{101} - 1)$ or 999....99 (100 nines). Therefore, A= $\frac{10^{100}\times 999...99}{100...01\times 999...99}$ (100 nines in $999...99$ )

By using the fact above, we can rewrite A as $\frac{999...99000...0}{9999...999}$ ) (999...9000...0 has 100 nines and 100 zeros; 9999...999 has 200 nines)

Hence, we can also rewrite A as $0.(999...99000...0)$ (999...9000...0 has 100 nines and 100 zeros)

Since we have to find 300 decimal digits of A, A = $0.(999...99000...0999...99)$ (999...99000...0999...99 has 100 nines then 100 zeros then 100 nines) So, the last 3 decimal digits of A are $\boxed{999}$