Null hypothesis, $H_0: p=0.9$
Alternate hypothesis, $H_A: p \neq 0.9$
Given, $\alpha=0.01$ and $n=225$
$$
\begin{gathered}
\alpha=P(|\bar{X}-p|>c \mid \mu=0.9) \\
\alpha=P\left(\left|\frac{\bar{X}-0.9}{\sqrt{\frac{0.9 \times 0.1}{225}}}\right|>\frac{c}{\sqrt{\frac{0.9 \times 0.1}{225}}}\right) \\
\alpha=P\left(|z|>\frac{c}{\sqrt{\frac{0.9 \times 0.1}{225}}}\right) \\
0.01=2 F_z\left(\frac{-15 c}{\sqrt{0.9 \times 0.1}}\right) \\
c=-\frac{\sqrt{0.9 \times 0.1}}{15} \times F_z^{-1}(0.005)=0.0516
\end{gathered}
$$
Since $|\bar{X}-p|=|0.894-0.9|<c, z$-test at significance level $(\alpha)=0.01$, will accept $H_0$.
Hence A is correct
