The probability of a man hitting a target is $$\frac{2}{5}$$. He fires at the target $$k$$ times ($$k$$, a given number). Then the minimum $$k$$, so that the probability of hitting the target at least once is more than $$\frac{7}{10}$$, is :

The probability of hitting the target in one shot is given as $$ \frac{2}{5} $$. Therefore, the probability of missing the target in one shot is $$ 1 - \frac{2}{5} = \frac{3}{5} $$.

The man fires $$ k $$ times independently. We need the probability that he hits the target at least once to be greater than $$ \frac{7}{10} $$. The event of hitting at least once is the complement of missing every shot. So, the probability of at least one hit is $$ 1 - \text{(probability of all misses)} $$.

The probability of missing all $$ k $$ shots is $$ \left( \frac{3}{5} \right)^k $$. Therefore, the probability of at least one hit is $$ 1 - \left( \frac{3}{5} \right)^k $$.

We set up the inequality:

$$ 1 - \left( \frac{3}{5} \right)^k > \frac{7}{10} $$

Subtract $$ \frac{7}{10} $$ from both sides:

$$ 1 - \frac{7}{10} > \left( \frac{3}{5} \right)^k $$

$$ \frac{3}{10} > \left( \frac{3}{5} \right)^k $$

Since both sides are positive and less than 1, we can take reciprocals and reverse the inequality sign:

$$ \frac{10}{3} < \left( \frac{5}{3} \right)^k $$

Now, $$ \frac{10}{3} \approx 3.333 $$ and $$ \frac{5}{3} \approx 1.6667 $$. We need the smallest integer $$ k $$ such that $$ \left( \frac{5}{3} \right)^k > \frac{10}{3} $$.

Check for $$ k = 1 $$:

$$ \left( \frac{5}{3} \right)^1 = \frac{5}{3} \approx 1.6667 $$

Since $$ 1.6667 < 3.333 $$, it does not satisfy.

Check for $$ k = 2 $$:

$$ \left( \frac{5}{3} \right)^2 = \frac{25}{9} \approx 2.7778 $$

Since $$ 2.7778 < 3.333 $$, it does not satisfy.

Check for $$ k = 3 $$:

$$ \left( \frac{5}{3} \right)^3 = \frac{125}{27} \approx 4.6296 $$

Since $$ 4.6296 > 3.333 $$, it satisfies the inequality.

Now, verify the original condition for $$ k = 3 $$:

Probability of all misses: $$ \left( \frac{3}{5} \right)^3 = \frac{27}{125} = 0.216 $$

Probability of at least one hit: $$ 1 - 0.216 = 0.784 $$

Since $$ 0.784 > 0.7 $$ (which is $$ \frac{7}{10} $$), it is correct.

Check if $$ k = 2 $$ works to ensure minimality:

Probability of all misses: $$ \left( \frac{3}{5} \right)^2 = \frac{9}{25} = 0.36 $$

Probability of at least one hit: $$ 1 - 0.36 = 0.64 $$

Since $$ 0.64 < 0.7 $$, it does not satisfy. Thus, $$ k = 3 $$ is the minimum.

The options are: A. 3, B. 5, C. 2, D. 4. Therefore, the correct option is A.

Hence, the correct answer is Option A.