Original concentration calculation homework
The mass of $\ce{FeCO3}$ is given as $m_{\ce{FeCO3}}=23.758\ \mathrm g$.

You already found that the molar mass of $\ce{FeCO3}$ is $M_{\ce{FeCO3}}=115.85\ \mathrm{g\ mol^{-1}}$.

For a pure sample, molar mass $M$ is defined as
$$M=\frac mn$$
where $m$ is mass and $n$ is amount of substance.
Thus, the amount of $\ce{FeCO3}$ can be calculated as follows:
$$\begin{align}n_{\ce{FeCO3}}&amp;=\frac {m_{\ce{FeCO3}}}{M_{\ce{FeCO3}}}\\[6pt]
&amp;=\frac {23.758\ \mathrm g}{115.85\ \mathrm{g\ mol^{-1}}}
\end{align}$$

The chemical equation
$$\ce{Fe^2+ + CO3^2- &lt;=&gt; FeCO3}$$
indicates that $1\ \mathrm{mol}$ of $\ce{Fe^2+}$ yields $1\ \mathrm{mol}$ of $\ce{FeCO3}$. Therefore, 
$$n_{\ce{Fe^2+}}=n_{\ce{FeCO3}}$$

The volume of the solution is given as $V=100.00\ \mathrm{ml}=0.10000\ \mathrm l$.

The concentration $c$ is defined as
$$c=\frac nV$$
where $n$ is amount of substance and $V$ is the volume of the solution.
Thus, the concentration of $\ce{Fe^2+}$ can be calculated as follows:
$$\begin{align}c_{\ce{Fe^2+}}&amp;=\frac {n_{\ce{Fe^2+}}}V\\[6pt]
&amp;=\frac {n_{\ce{Fe^2+}}}{0.10000\ \mathrm l}\end{align}$$