Find sample size given the number of desired successes I am revising for an upcoming statistics test and found a question that I'm struggling to understand the answer to.
Before the question, you should know that to answer the question, the below formula should be used:
$$n=\left(\frac{2\space\times\space1.96}{w}\right)^2\times \hat p\times(\hat p-1)$$
Where $n$ is the sample size, $w$ is the width, and $\hat p$ is the sample proportion. This is only high school level maths, and this is the equation that should be used.
The question itself is:

Monique is practicing gosling for netball. She knows from past experience that her chance of making any one shot is about 70%. Her coach has asked her to keep practising until she scores 50 goals. How many shots would she need to attempt to ensure that the probability of scoring at least 50 goals is 0.99?

From the question, I know that $\hat p \space = 0.7$. I'm assuming that this is a 95% confidence interval, and so that's why I used 1.96 in the formula for $n$ (1.96 standard deviations).
However, I can't find a way to find $w$. All previous questions have explicitly given $w$ along with the confidence interval.
The answer given is 86, but the steps taken to achieve this result were not given. So, in the above question, how can the $w$ value be found?
 A: The random number of successes $X$ in $n$ trials with a probability $p = 0.7$ of success on any given trial is $$X \sim \operatorname{Binomial}(n, p = 0.7),$$ and we want to find the smallest $n$ such that $\Pr[X \ge 50] \ge 0.99$.  Using a normal approximation to the binomial, we have $$\Pr\left[ \frac{X - np}{\sqrt{np(1-p)}} \ge \frac{50 - (0.7)n}{\sqrt{(0.21)n}}\right] \approx \Pr\left[Z \ge \frac{50 - (0.7)n}{\sqrt{(0.21)n}}\right] \ge 0.99,$$ and since the $99^{\rm th}$ percentile of the standard normal distribution is $2.32635$, we require $$\frac{50 - (0.7)n}{\sqrt{(0.21)n}} \le -2.32635.$$  Solving this inequality yields $n \ge 85.5117$, or $n \ge 86$ since it must be an integer.
To check the approximation gives a tight lower bound, with a computer or Wolfram Alpha we can check that $$\Pr[X \le 49 \mid n = 85] = \sum_{x=0}^{49} \binom{85}{x} (0.7)^x (0.3)^{85-x} \approx 0.0104508 > 0.01,$$ whereas $$\Pr[X \le 49 \mid n = 85] = \sum_{x=0}^{49} \binom{86}{x} (0.7)^x (0.3)^{86-x} \approx 0.00709005 < 0.01,$$ so $n = 86$ is indeed correct.

So, where does your formula $$n = \left(\frac{2 \times 1.96}{w}\right)^2 \hat p ( 1 - \hat p)$$ come from and what does it have to do with your question?
The issue here is that $1.96$ is the critical value for a two-sided $95\%$ confidence interval.  It doesn't have anything to do with guaranteeing a $99\%$ probability of success.  And $\hat p$ is an estimate of $p$, but in our case here, we know that $p = 0.7$ because that is the historical performance we have been given.  In fact, the question really has nothing to do with confidence intervals, as I have already shown.
