It is given that the probability, a missile hits a target successfully $$p = {3 \over 4}$$, so the probability to not hits the target is $${1 \over 4}$$. And it is also given that to destroy the target completely, at least three successful hits are required.

Now, according to the question, let the minimum number of missiles required to fired is n, so

$$^n{C_3}{p^3}{q^{n - 3}}{ + ^n}{C_4}{p^4}{q^{n - 4}} + ...{ + ^n}{C_n}{p^n}\, \ge \,0.95$$

$$ \Rightarrow 1 - \left\{ {^n{C_0}{{\left( {{1 \over 4}} \right)}^n}{ + ^n}{C_1}\left( {{3 \over 4}} \right){{\left( {{1 \over 4}} \right)}^{n - 1}}{ + ^n}{C_2}{{\left( {{3 \over 4}} \right)}^2}{{\left( {{1 \over 4}} \right)}^{n - 2}}} \right\}\, \ge \,0.95$$

$$ \Rightarrow 1 - {{95} \over {100}}\, \ge \,{1 \over {{4^n}}} + {{3n} \over {{4^n}}} + {{n(n - 1)} \over 2}{9 \over {{4^n}}}$$

$$ \Rightarrow {{{4^n}} \over {20}}\, \ge \,{{2 + 6n + 9{n^2} - 9n} \over 2}$$

$$ \Rightarrow 10(9{n^2} - 3n + 2)\, \le \,{4^n}$$

Now, at n = 3, LHS = 720, RHS = 64

at n = 4, LHS = 1340, RHS = 256

at n = 5, LHS = 2120, RHS = 1024

at n = 6, LHS = 3080, RHS = 4096

Hence, n = 6 missiles should be fired.