How much water is required to dilute glacial acetic acid to given pH?
The question is flawed; the $\mathrm{pH}$ of 100 % glacial acetic acid is not $\mathrm{pH=2.4}$. (Note that 100 % glacial acetic acid contains no water; therefore you cannot apply the usual equations for dilute aqueous solutions to calculate the $\mathrm{pH}$ of 100 % glacial acetic acid.)

Also note, since acid dissociation is not a redox reaction, you cannot apply the Nernst equation here.

The quantity $\mathrm{pH}$ is actually defined in terms of the activity of $\ce{H+}$ ions; however, for simplicity’s sake, we may use the old concept of $\mathrm{pH}$ with the definition in terms of the concentration of $\ce{H+}$ ions
$$\mathrm{pH}=-\log_{10}[\ce{H+}]$$
or
$$[\ce{H+}]=10^{-\mathrm{pH}}$$

The equilibrium of acid dissociation is usually written as:
$$\ce{HA &lt;=&gt; A- + H+}$$
where $\ce{HA}$ is a generic acid that dissociates into the conjugate base $\ce{A-}$.
The equilibrium constant $K_\mathrm a$ for this reaction is defined as
$$K_\mathrm a=\frac{[\ce{H+}]\cdot [\ce{A-}]}{[\ce{HA}]}$$
Usually, values are given for the logarithmic constant $\mathrm pK_\mathrm a$, which is defined as
$$\mathrm pK_\mathrm a=-\log_{10}K_\mathrm a$$
or
$$K_\mathrm a=10^{-\mathrm pK_\mathrm a}$$

The corresponding equilibrium for acetic acid is
$$\ce{CH3COOH &lt;=&gt; CH3COO- + H+}$$
The value of the logarithmic constant for acetic acid at 25 °C is
$$\mathrm pK_\mathrm a=4.756$$
Thus, the value for the acid dissociation constant is
$$K_\mathrm a=\frac{[\ce{H+}]\cdot [\ce{CH3COO-}]}{[\ce{CH3COOH}]}=10^{-\mathrm pK_\mathrm a}=10^{-4.756}\approx1.75\times10^{-5}$$

The chemical equation shows that, one $\ce{CH3COO-}$ ion and one $\ce{H+}$ ion are produced for every $\ce{CH3COOH}$ molecule that dissociates. This can be represented by subtracting a value $x$ from the original acetic acid concentration $[\ce{CH3COO-}]_0$, and adding this value $x$ to the concentrations of $\ce{CH3COO-}$ and $\ce{H+}$, i.e. when neglecting the $\ce{H+}$ ions from the autoprotolysis of water:

$$\begin{align}
[\ce{CH3COOH}]&amp;=[\ce{CH3COOH}]_0-x\\[6pt]
[\ce{CH3COO-}]&amp;=x\\[6pt]
[\ce{H+}]&amp;=x
\end{align}$$

Thus, the equation for the acid dissociation equilibrium can be rewritten and solved for the original acetic acid concentration $[\ce{CH3COOH}]_0$ as follows:

$$\begin{align}
K_\mathrm a&amp;=\frac{[\ce{H+}]\cdot [\ce{CH3COO-}]} {[\ce{CH3COOH}]}\\[6pt]
&amp;=\frac{x\cdot x}{[\ce{CH3COOH}]_0-x}\\[6pt]
&amp;=\frac{x^2}{[\ce{CH3COOH}]_0-x}\\[6pt]
K_\mathrm a\cdot\left([\ce{CH3COOH}]_0-x\right)&amp;=x^2\\[6pt]
[\ce{CH3COOH}]_0-x&amp;=\frac{x^2}{K_\mathrm a}\\[6pt]
[\ce{CH3COOH}]_0&amp;=\frac{x^2}{K_\mathrm a}+x
\end{align}$$

Since $[\ce{H+}]=x$, $[\ce{H+}]=10^{-\mathrm{pH}}$, and $K_\mathrm a=10^{-\mathrm pK_\mathrm a}$, the required acetic acid concentration $[\ce{CH3COOH}]_0$ can be given in terms of the desired $\mathrm{pH}$:

$$\begin{align}
[\ce{CH3COOH}]_0&amp;=\frac{x^2}{K_\mathrm a}+x\\[6pt]
&amp;=\frac{[\ce{H+}]^2}{K_\mathrm a}+[\ce{H+}]\\[6pt]
&amp;=\frac{\left(10^{-\mathrm{pH}}\right)^2}{K_\mathrm a}+10^{-\mathrm{pH}}\\[6pt]
&amp;=\frac{\left(10^{-\mathrm{pH}}\right)^2}{10^{-\mathrm pK_\mathrm a}}+10^{-\mathrm{pH}}\\[6pt]
&amp;=10^{\mathrm pK_\mathrm a-2\mathrm{pH}}+10^{-\mathrm{pH}}
\end{align}$$

For example, for the given acetic acid solution with $\mathrm{pH}=2.4$:

$$\begin{align}
[\ce{CH3COOH}]_0&amp;=10^{\mathrm pK_\mathrm a-2\mathrm{pH}}+10^{-\mathrm{pH}}\\[6pt]
[\ce{CH3COOH}]_0&amp;=10^{4.756-2\times2.4}+10^{-2.4}\\[6pt]
&amp;=0.907630545\\[6pt]
&amp;\approx0.9
\end{align}$$

and for the desired acetic acid solution with $\mathrm{pH}=4.3$:

$$\begin{align}
[\ce{CH3COOH}]_0&amp;=10^{\mathrm pK_\mathrm a-2\mathrm{pH}}+10^{-\mathrm{pH}}\\[6pt]
[\ce{CH3COOH}]_0&amp;=10^{4.756-2\times4.3}+10^{-4.3}\\[6pt]
&amp;=0.000193338\\[6pt]
&amp;\approx0.0002
\end{align}$$

Finally, since 
$$c_1\cdot V_1=c_2\cdot V_2$$
you may use the concentrations to calculate how much the given acetic acid solution with $\mathrm{pH}=2.4$ should be diluted in order to obtain the desired acetic acid solution with $\mathrm{pH}=4.3$.