The formula
$$c_1\cdot V_1=c_2\cdot V_2$$
is correct.
Since your stock solution has a concentration of $c=0.1\ \mathrm{mol\ l^{-1}}$, the value for the initial concentration $c_1$ is always
$$c_1=0.1\ \mathrm{mol\ l^{-1}}$$
Since you want the final volume of each solution to be $V=30\ \mathrm{ml}$, the value for the final volume $V_2$ is always
$$V_2=30\ \mathrm{ml}$$
You want four individual solutions (labelled $\mathrm a$, $\mathrm b$, $\mathrm c$, and $\mathrm d$) with four different final concentrations ($0.1\ \mathrm{mol\ l^{-1}}$, $0.05\ \mathrm{mol\ l^{-1}}$, $0.025\ \mathrm{mol\ l^{-1}}$, $0.0125\ \mathrm{mol\ l^{-1}}$. These are the individual values for the final concentration $c_2$:
$$\begin{align}
c_{2,\mathrm{a}}&=0.1\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{b}}&=0.05\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{c}}&=0.025\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{d}}&=0.0125\ \mathrm{mol\ l^{-1}}
\end{align}$$
Now you can rearrange the above-mentioned equation to solve for the initial volume of the stock solution $V_1$:
$$\begin{alignat}{2}
V_1&=\frac{c_2\cdot V_2}{c_1}\\[6pt]
V_{1,\mathrm a}&=\frac{0.1\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=30\ \mathrm{ml}\\[6pt]
V_{1,\mathrm b}&=\frac{0.05\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=15\ \mathrm{ml}\\[6pt]
V_{1,\mathrm c}&=\frac{0.025\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=7.5\ \mathrm{ml}\\[6pt]
V_{1,\mathrm d}&=\frac{0.0125\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=3.75\ \mathrm{ml}
\end{alignat}$$
You measure these volumes of the stock solution and transfer them into four separate vessels.
Finally, you have to fill up each solution with pure water to the final volume of $V_2=30\ \mathrm{ml}$. The required additional volume of water $V_{\ce{H2O}}$ is approximately
$$\begin{alignat}{2}
V_{\ce{H2O}}&\approx V_2-V_1\\[6pt]
V_{\ce{H2O},\mathrm a}&\approx30\ \mathrm{ml}-30\ \mathrm{ml}&&=0\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm b}&\approx30\ \mathrm{ml}-15\ \mathrm{ml}&&=15\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm c}&\approx30\ \mathrm{ml}-7.5\ \mathrm{ml}&&=22.5\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm d}&\approx30\ \mathrm{ml}-3.75\ \mathrm{ml}&&=26.25\ \mathrm{ml}
\end{alignat}$$
