Upon testing 80 resistors manufactured by a certain company, it is found that 15 resistors failed to meet the tolerance design specifications
a) Construct a 92% two-sided confidence interval for the true fraction of resistors manufactured by the company that do meet the tolerance design specifications.
The $1-\alpha$ confidence interval for $\hat p$ is
$$\large{\left[ \hat p-z_{(1-\frac{\alpha}{2})} \cdot \sqrt{\frac{\hat p(1-\hat p)}{ n}} , \hat p+z_{(1-\frac{\alpha}{2})} \cdot \sqrt{\frac{\hat p(1-\hat p)}{ n}} \right]}$$
$z_{(1-\frac{\alpha}{2})}$ is the z-value of the standard normal distribution. $1-\alpha$ is the confidence level.
In your case $\hat p=0.65$ This is right.
$p(1-\hat p)=0.65\cdot 0.35$ and $1-\frac{\alpha}{2}=1-\frac{0.08}{2}=0.96$. Finally you know that $n=80$.
And $z_{0.96}=1.75$. Just look at the table for the standard normal distribution.
I think with all the given values you can evaluate the bounds of the confidence interval.