This is a classic case of hypothesis testing by binomial distribution. You now follow the recipe above to answer the task and select $5$ % level of significance since it is not a question of medication for a serious illness.

1.

The null hypothesis is that the new drug is worse or as good as the old one:

$$p=0.85$$

The alternative hypothesis in this case is that the new drug is better. The reason for this is that you only need to know if you are going to approve for sale and thus the new drug must be better:

$$p>0.85$$

2.

You see that the event occurs 92 times, since the new drug cured 92 of 103 patients.

3.

You then calculate the probability of seeing at least one equally high observation in a digital probability distribution tool:

$$\begin{array}{llll}\hfill P\left(X\ge 92\right)& =1-P\left(X\le 91\right)& \hfill & \\ \hfill & =0.136& \hfill & \end{array}$$

$$\begin{array}{lll}\hfill P\left(X\ge 92\right)=1-P\left(X\le 91\right)=0.136& & \hfill \end{array}$$

This result indicates that there is a $13.6$ % chance that more than 92 patients would be cured with the old medicine.

4.

You have decided that if the old medicine is to be replaced, then your $p$ value must be less than $5$ %. Therefore, there must be less than a $5$ % chance of rejecting $H_0$ when the new drug is not better than the old drug. Your $p$ value is

$$p=13.6\text{\%}>5\text{\%}$$

so $H_0$ cannot be rejected. The new drug does not enter the market.

If the $p$ value had been less than the level of significance, that would mean that the new drug represented by the alternative hypothesis is better, and that you are sure of this with statistical significance.