Wald statistic has form $$ Z_W = (\hat{p} - p_0) \sqrt{\mathcal{I(\hat{p})}} \sim N(0,1). $$
After substitution $\hat{p} = \frac{x}{N}$ and $I(\hat{p}) = \frac{N^3}{X (N - X)}$ we get $$ Z_W = (\frac{X}{N} - p_0) \sqrt{\frac{N^3}{X (N - X)}}. $$
Note that since $X$ is still a random variable, so $Z_W$ is a random variable too and $Z_W \sim N(0,1)$. Let's get its realization using our data $x=534$ and $N=1000$ $$ z_W = (\frac{x}{N} - p_0) \sqrt{\frac{N^3}{x (N - x)}} = (0.534 - 0.5) \sqrt{\frac{1000^3}{534\cdot466}} = 2.155 $$
Critical region of $N(0,1)$ is $$ \mathcal{W} = (-\infty, u_{1-\alpha/2}) \cup (u_{\alpha/2}, \infty) = (-\infty, -1.96) \cup (1.96, \infty), $$ where $u_p$ is $1 - p$ quantile of standard normal distribution $N(0,1)$.
Wald confidence interval $$ (d_W, h_W) = \left(\hat{p} - u_{\alpha/2} \frac{1}{\sqrt{\mathcal{I(\hat{p})}}}, \hat{p} + u_{\alpha/2} \frac{1}{\sqrt{\mathcal{I(\hat{p})}}} \right) $$ after substitution $$ (d_W, h_W) = \left(\frac{x}{N} - u_{\alpha/2} \sqrt{\frac{x (N - x)}{N^3}}, \frac{x}{N} + u_{\alpha/2} \sqrt{\frac{x (N - x)}{N^3}} \right) $$ using $x=534$ and $N = 1000$ $$ (d_W, h_W) = \left(0.534 - 1.96 \cdot 0.0158, 0.534 + 1.96 \cdot 0.0158 \right) = (0.503, 0.565) $$