Answer: Option B
The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101.
The next number that will leave a remainder of 2 when divided by 3 is 104, 107, ....
The largest 3 digit number that will leave a remainder of 2 when divided by 3 is 998.
So, it is an AP with the first term being 101 and the last term being 998 and common difference being 3.
Sum of an AP =
$$\left( {\frac{{{\text{First Term}} + {\text{Last Term}}}}{2}} \right) \times $$     $${\text{Number}}\,{\text{of}}\,{\text{Terms}}$$
We know that in an A.P., the nth term an = a1 + (n - 1) × d
In this case, therefore, 998 = 101 + (n - 1) × 3
i.e., 897 = (n - 1) × 3
Therefore, n - 1 = 299
Or n = 300
Sum of the AP will therefore, be
$$\eqalign{
& = \frac{{101 + 998}}{2} \times 300 \cr
& = 1,64,850 \cr} $$