-    By hit and trial, we find that a number exactly divisible by 7 and consisting entirely of nines is 999999.
 
  Number of digits in it = 6

Let the required number be x.

When x is multiplied by 7, the result consists entirely of nines, which means the result is of the form 999...999 (nines repeated k times, where k is the number of digits in the result).

Let the number of digits in x be n. Then we can write x as:

x = (999...999)/7 = (10^n - 1)/7

We need to find the smallest n such that x is a whole number.

For x to be a whole number, 10^n - 1 must be divisible by 7.

Now, we can use Fermat's Little Theorem which states that if p is a prime number, then for any integer a not divisible by p, a^(p-1) - 1 is divisible by p.

Here, p=7 is a prime number and a=10. So, we have:

10^(7-1) - 1 = 10^6 - 1 is divisible by 7.

Hence, n must be a multiple of 6 for x to be a whole number.

Also, since we want to find the smallest n, we can try n=6 and check if x is a whole number.

x = (10^6 - 1)/7 = 142857

We can verify that x multiplied by 7 gives 999999.

Therefore, the smallest number that satisfies the given condition has 6 digits.

Answer: (C) 6.