This is indeed a question about the inverse of a normal cumulative distribution function (CDF). Such inverse functions are often called quantile functions. The particulars of the computation depend on whether you are using software or printed normal tables.

Software: If you have software available, you can use your equation $P(X \le C) = 0.22,$ where $X \sim \mathsf{Norm}(\mu = 4,\, \sigma = 7).$ In R statistical software normal quantile functions are denoted qnorm, so the answer $-1.405$ can be obtained as:

qnorm(.22, 4, 7)
# [1] -1.405352

This is similar to @Henry's code, in which the argument lower.tail=F takes 22% of the probability from the upper tail, whereas qnorm without this argument takes 22% from the lower tail (according to the usual definition of the CDF).

Other statistical and mathematical software and calculators perform similar computations using different syntax.

The plot below shows the PDF and CDF of $\mathsf{Norm}(4, 7)$ with red dashed lines indicating key quantities.

Printed tables: If you doing this with printed standard normal CDF tables, then you need to 'standardize' much as you have done:

$$P(X \le C) = P\left(\frac{X - \mu}{\sigma} \le \frac{C - 4}{7}\right) = P\left(Z \le \frac{C-4}{7}\right) = 0.22,$$ where $Z \sim \mathsf{Norm}(0,1).$ Looking in the body of the table you can find a z-score that corresponds as nearly as possible to 0.22 (where the meaning of 'corresponds' depends on the format of the table). Here, the corresponding 'z-score' in the margin of the table is $-0.77.$ (Most tables allow no more than two-place accuracy).

Then $(C - 4)/7 = -0.77)$ so that $C \approx -1.39.$ But you can trust only two significant digits of the answer, so it amounts to $-1.4.$ By interpolation in the body of the table, you might get a little more accuracy. But your answer is about as close as you can reliably get using printed normal tables.

To calculate the inverse you will need to know the mean and standard deviation of the original x, as there are there are many values of the input which can result in the same z scores. As I think you've realised the mean and standard deviaiton of the z-scores are always 0 and 1 respectively, by the definition of the z-score and so give no useful information.

If you know the original mean and standard deviation. You can calculate the original x simply as x=std(x)*z+mean(x). However, if you don't then the problem is not well defined and can't be solved.

As an example of this if you add 100 to all your original data, the z-scores are unchanged. There is not way of distinguishing which x value is correct from the z-score alone. The same if you multiply the original data by a constant.