The binomial distribution applies to a finite and fixed number of independent Bernoulli trials, each with the same probability of "success" $p$. Under these criteria, a binomially distributed random variable $X$ then counts the total number of successes observed out of $n$ trials. As such, it does not model your stated event.
What you want is a negative binomial random variable, one parametrization of which counts the total number of trials needed to obtain $r$ successes. You can see how this differs from the regular binomial scenario, which counts the successes when the number of trials is fixed: in the negative binomial case, we count the total number of trials when the number of successes is fixed.
In your case, the number of successes is fixed at $r = 3$, and we wish to find the probability that it takes $X = 7$ trials to observe three successes (where "success" is defined here as observing a defective product). The probability distribution for $X$ is given by $$\Pr[X = x] = \binom{x - 1}{x - r} p^r (1-p)^{x-r}, \quad x = r, r+1, r+2, \ldots.$$ So for $x = 7$, $r = 3$, $p = 0.05$, you will get your desired probability of observing the third defect on the seventh trial.
Where does this formula come from? We reason as follows. Clearly, in order for the $r^{\rm th}$ success to occur on the $x^{\rm th}$ trial, the last trial must be successful and that's when we stop sampling. Now there are $r-1$ successes to be distributed among the $x-1$ trials and there are exactly $\dbinom{x-1}{r-1} = \dbinom{x-1}{x-r}$ ways to do this. There are $r$ successes that occur with probability $p$ and $x-r$ failures that occur with probability $1-p$, so the probability of the desired outcome is what is claimed above.
