A man saves Rs. 200 in each of the first three months of his service. In each of the subsequent months his saving increases by Rs. 40 more than the saving of immediately previous month. His total saving from the start of service will be Rs. 11040 after:

For the first three months the man saves a fixed amount of Rs. 200 every month.

Month-wise saving pattern:
  Month 1 = Rs. 200
  Month 2 = Rs. 200
  Month 3 = Rs. 200
From the 4th month onward each month’s saving is Rs. 40 more than the previous month, so an arithmetic progression (A.P.) begins with

First term of the A.P.: $$a_1 = 240$$ (saving in the 4th month)
Common difference: $$d = 40$$

Let the total number of months be $$n$$. For $$n \gt 3$$, the A.P. part has $$m = n - 3$$ terms.

Sum of the first three constant-saving months:
$$S_{\text{first 3}} = 3 \times 200 = 600$$

Sum of the A.P. for the next $$m$$ months:
Using $$S_m = \frac{m}{2}\bigl(2a_1 + (m-1)d\bigr)$$, we get
$$S_{\text{A.P.}} = \frac{m}{2}\bigl(2 \times 240 + (m-1)\,40\bigr)$$
$$= \frac{m}{2}\bigl(480 + 40m - 40\bigr)$$
$$= \frac{m}{2}\bigl(440 + 40m\bigr)$$
$$= 20m\,(11 + m)$$

Total saving after $$n$$ months:
$$S_{\text{total}} = 600 + 20m(11 + m)$$

Given $$S_{\text{total}} = 11040$$, substitute and solve for $$m$$:

$$600 + 20m(11 + m) = 11040$$
$$20m(11 + m) = 10440$$
Divide by 20:
$$m(m + 11) = 522$$
$$m^2 + 11m - 522 = 0$$

Quadratic formula:
Discriminant $$D = 11^2 + 4 \times 522 = 121 + 2088 = 2209 = 47^2$$
$$m = \frac{-11 \pm 47}{2}$$

Positive root: $$m = \frac{-11 + 47}{2} = \frac{36}{2} = 18$$ (negative root is inadmissible).

Therefore the total months are
$$n = m + 3 = 18 + 3 = 21$$

Hence the man will accumulate Rs. 11040 in 21 months.

Option C which is: 21 months