Solution:
Let us first express $a$ and $b$ in terms of their digits.

$a$ consists of $2010$ digits, all $4$'s. So,
$$a = \underbrace{444\cdots 444}_{2010\ \text{digits}} = 4 \times \underbrace{111\cdots 111}_{2010\ \text{digits}}$$
Let $N = \underbrace{111\cdots 111}_{2010\ \text{digits}}$. Then $a = 4N$.

Similarly,
$$b = \underbrace{999\cdots 999}_{2010\ \text{digits}} = 10^{2010} - 1$$

Now, the product:
$$a b = (4N)(10^{2010} - 1) = 4N \cdot 10^{2010} - 4N$$

Let us compute $N$:
$$N = 111\cdots 111 = \frac{10^{2010} - 1}{9}$$
So,
$$4N = \frac{4(10^{2010} - 1)}{9}$$

Therefore,
$$a b = 4N \cdot 10^{2010} - 4N = 4N (10^{2010} - 1) = 4N \cdot 10^{2010} - 4N$$

Let us write $4N$ explicitly:
$4N$ is a number with $2010$ digits, all $4$'s, so $444\cdots 444$ ($2010$ digits).

So $4N \cdot 10^{2010}$ is the number $444\cdots 444$ ($2010$ digits) followed by $2010$ zeros.

Subtract $4N$:
$$a b = 444\cdots 444\underbrace{000\cdots 000}_{2010\ \text{zeros}} - 444\cdots 444$$

This is the same as writing:
$$\left(\begin{array}{c} \text{444...444} \\ \text{(2010 digits)} \\ \text{000...000} \\ \text{(2010 zeros)} \end{array}\right) - \left(\begin{array}{c} \text{000...000} \\ \text{444...444} \end{array}\right)$$

So, the subtraction is:
- The first $2010$ digits are $444\cdots 444$ minus $0$'s, so $444\cdots 444$.
- The last $2010$ digits are $0$'s minus $444\cdots 444$.

But since we are subtracting $444\cdots 444$ from $000\cdots 000$, we need to borrow.

Let us perform the subtraction:

Let $X = 444\cdots 444\ 000\cdots 000$ ($2010$ $4$'s, $2010$ $0$'s), and subtract $Y = 000\cdots 000\ 444\cdots 444$ ($2010$ $0$'s, $2010$ $4$'s).

So, the result is:
- The first $2010$ digits: $444\cdots 444$ minus $0$'s = $444\cdots 444$.
- The last $2010$ digits: $0$'s minus $444\cdots 444$ = need to borrow.

Let us look at the last $2010$ digits:

Subtracting $444\cdots 444$ from $000\cdots 000$ means that we get $999\cdots 999 - 444\cdots 444 + 1$ (since $10^k - x = 999\cdots 999 - x + 1$ for $k$ digits).

So, the last $2010$ digits are $999\cdots 999 - 444\cdots 444 + 1 = 555\cdots 555 + 1 = 555\cdots 556$ ($2010$ digits, last digit $6$).

But, since we borrowed $1$ from the previous digit, the $2010$th digit from the left is the last digit of the first $2010$ digits, after borrowing $1$.

Let us clarify:
- The number has $2010 + 2010 = 4020$ digits (possibly $4019$ if there is a leading zero, but in this case, the subtraction does not reduce the number of digits).
- The first $2010$ digits are $444\cdots 444$ minus $1$ (because of the borrow).

So, $444\cdots 444 - 1 = 444\cdots 443$ (the last digit is $3$).

Therefore, the first $2009$ digits are $4$'s, and the $2010$th digit is $3$.

Thus, the $2010$th digit of $ab$ is $3$.