To determine the value of $$\mathrm{m}$$, we need to formulate the problem using some basic concepts of arithmetic progression and work. Let's first understand the nature of the problem:

Initially, there are $$\mathrm{m}$$ computers, and it is estimated that with these $$\mathrm{m}$$ computers, the assignment can be completed in 17 days.

However, due to the crash of 4 computers every day starting from the second day onward, the total time taken extends by 8 days, making it 25 days in total.

To begin with, let's define the total work (W) in terms of the number of computers and days:

The total work (W) is given by:

The amount of work completed each day with $$\mathrm{m}$$ computers for 17 days:

$$W = 17m$$

When computers crash, the number of working computers each day forms an arithmetic sequence. On the first day, there are $$\mathrm{m}$$ computers. On the second day, there are $$\mathrm{m} - 4$$ computers, on the third day, there are $$\mathrm{m} - 8$$ computers, and so on. We need to sum this series until 25 days are completed.

This can be formulated as:

Total work done over 25 days with decrement in the number of computers:

$$W = m + (m - 4) + (m - 8) + \ldots + \left[m - 4 \times (n - 1)\right]$$

where $$n$$ is the number of days. Here, $$n = 25$$.

Notice that we form an arithmetic series where the first term (a) is $$\mathrm{m}$$ and the common difference (d) is -4. The sum of the first n terms of an arithmetic series is:

$$S_n = \frac{n}{2} \left[ 2a + (n - 1)d \right]$$

Plugging in the values:

$$S_{25} = \frac{25}{2} \left[ 2m + (25 - 1)(-4) \right]$$

$$S_{25} = \frac{25}{2} \left[ 2m - 96 \right]$$

$$S_{25} = \frac{25}{2} \left[ 2m - 96 \right] = 25(m - 48)$$

This work should be equivalent to the work calculated earlier, so:

$$17m = 25(m - 48)$$

Solving for $$\mathrm{m}$$:

$$17m = 25m - 1200$$

$$8m = 1200$$

$$m = 150$$

Thus, the value of $$\mathrm{m}$$ is equal to:

Option C: 150