Mass of chromium in ammonium dichromate
To find the mass of chromium, one needs to know the amount $n$ ($m = nM$). Each formula unit of ammonium dichromate $\ce{(NH4)2Cr2O7}$ contains $2$ $\ce{Cr}$, therefore

$$n(\ce{Cr}) = 2 \cdot n(\ce{(NH4)2Cr2O7})$$

The amount of ammonium dichromate can be found from its concentration (as you correctly did):

$$n(\ce{(NH4)2Cr2O7}) = C(\ce{(NH4)2Cr2O7}) \cdot V$$

Putting it all together, the final expression for the mass of chromium is

\begin{align}
m(\ce{Cr}) &amp;= 2 \cdot C(\ce{(NH4)2Cr2O7}) \cdot V \cdot M(\ce{Cr}) \\
&amp;= 2 \cdot \pu{0.04 mol L-1} \cdot \pu{0.1 L} \cdot \pu{51.996 g mol-1} \\
&amp;= \pu{0.416 g}
\end{align}

The answer deviates with the one provided by the textbook; however, if you were asked to find the mass of chromium(VI) oxide $\ce{CrO3}$ instead, it fits perfectly:

\begin{align}
m(\ce{CrO3}) &amp;= 2 \cdot C(\ce{(NH4)2Cr2O7}) \cdot V \cdot M(\ce{CrO3}) \\
&amp;= 2 \cdot \pu{0.04 mol L-1} \cdot \pu{0.1 L} \cdot \pu{99.99 g mol-1} \\
&amp;= \pu{0.80 g}
\end{align}

I deduced $\ce{CrO3}$ by backtracking to the molecular weight of the compound, which should be around $\pu{100 g mol-1}$, and among common chromium substances this is the only one that comes to mind.