To determine the time required for a certain level of completion ($x\%$) of a first-order reaction, we can use the formula that relates the time $ t $, the rate constant $ k $, and the concentration of the reactant. The formula for a first-order reaction, when expressed in terms of the initial concentration $[A]_0$ and the concentration at time $t$, $[A]_t$, is given by the integrated rate law for first-order reactions:

$$\ln\left(\frac{[A]_0}{[A]_t}\right) = kt$$

For a given percentage of completion, we use the fact that $[A]_t = [A]_0(1-\frac{x}{100})$, where $x$ is the percentage completion. Thus, the equation becomes:

$$\ln\left(\frac{[A]_0}{[A]_0(1-\frac{x}{100})}\right) = kt$$

Simplifying, we get:

$$\ln\left(\frac{1}{1-\frac{x}{100}}\right) = kt$$

Let's calculate the time required for $99.9\%$ and $90\%$:

For $99.9\%$ completion ($x = 99.9\%$):

$$\ln\left(\frac{1}{1-\frac{99.9}{100}}\right) = kt_{99.9}$$

$$\ln\left(\frac{1}{0.001}\right) = kt_{99.9}$$

$$\ln(1000) = kt_{99.9}$$

For $90\%$ completion ($x = 90\%$):

$$\ln\left(\frac{1}{1-\frac{90}{100}}\right) = kt_{90}$$

$$\ln\left(\frac{1}{0.1}\right) = kt_{90}$$

$$\ln(10) = kt_{90}$$

Since $k$ is a constant for a given reaction at a constant temperature, we can compare the times $t_{99.9}$ and $t_{90}$ directly by comparing the logarithmic values:

$$\frac{t_{99.9}}{t_{90}} = \frac{\ln(1000)}{\ln(10)}$$

Using the property of logarithms, we know that $\ln(1000) = 3\ln(10)$ because $1000 = 10^3$, so:

$$\frac{t_{99.9}}{t_{90}} = \frac{3\ln(10)}{\ln(10)} = 3$$

Therefore, the time required for $99.9\%$ completion of a first-order reaction is $3$ times the time required for completion of $90\%$ of the reaction. The nearest integer to this value is $3$.