How to determine the volume of water necessary to produce a solution with 4.9 g/l sulfuric acid?
We have talked about it in chat and maybe the question is a bit ambiguous. I would interpret the last line as a mass concentration. So what you are actually looking to create is a $\gamma(\ce{H2SO4}) = \pu{4.9 g/l}$ solution. Then the calculation becomes straight forward.
Keep in mind that water is your solvent, so you can't actually dilute it; also writing down the concentration is a bit pointless here.

You can easily convert the mass concentration to an amount concentration. The relative molar mass $M$ is given by the mass $m$ of the substance divided by the amount of substance $n$:
$$M = \frac{m}{n}\tag1\label{molarmass}$$

The mass concentration is defined as 
$$\gamma = \frac{m}{V}.\tag2\label{massconc}$$
The amount concentration is defined as 
$$c = \frac{n}{V}.\tag3\label{amountconc}$$
You can use \eqref{molarmass} in \eqref{amountconc} and obtain
$$c = \frac{m}{M} \times \frac{1}{V}
=\frac{m}{V} \times \frac{1}{M}.\tag4$$
Substitute \eqref{massconc} back in and you have
$$c = \frac{\gamma}{M}.\tag5\label{amounttomass}$$

Now in the question you have given the following:
\begin{align}
M(\ce{H2SO4}) &amp;= \pu{98.1 g/mol}\\
c_\mathrm{i}(\ce{H2SO4}) &amp;= \pu{0.1 mol/l}\\
V_\mathrm{i}(\ce{H2SO4}) &amp;= \pu{100 ml}\\
\gamma_\mathrm{f}(\ce{H2SO4}) &amp;= \pu{4.9 g/l}
\end{align}
I have given the initial values the subscript $\mathrm{i}$ and the final values the subscript $\mathrm{f}$ (instead of concentrated and diluted).

You are now looking for the volume of solvent you have to add, which is 
$$\Delta V = V_\mathrm{f} - V_\mathrm{i}.\tag5\label{volume}$$

Then you are right, you need the following relation
$$c_\mathrm{i} \times V_\mathrm{i} = c_\mathrm{f} \times V_\mathrm{f},
\tag6$$
rearranged and using \eqref{amounttomass}
\begin{align}
c_\mathrm{i} \times V_\mathrm{i} &amp;=
\frac{\gamma_\mathrm{f}}{M} \times V_\mathrm{f},\\
V_\mathrm{f} &amp;=
\frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i},
\end{align}
and using \eqref{volume}
\begin{align}
V_\mathrm{i} + \Delta V &amp;=
\frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i},\\
\Delta V &amp;=
\frac{M}{\gamma_\mathrm{f}} \times c_\mathrm{i} \times V_\mathrm{i} - V_\mathrm{i}\\
&amp;= \left(\frac{M \times c_\mathrm{i}}{\gamma_\mathrm{f}} - 1\right)V_\mathrm{i}.
\tag7\label{dilute}
\end{align}

Now simply plug in your values and get the solution (hover over box to reveal):


   $\displaystyle \Delta V = \left(\frac{\pu{98.1 g//mol} \times \pu{0.1 mol//l}}{\pu{4.9 g//l}} - 1\right)\times \pu{0.100 l} = \pu{0.100 l}.$