Determining water of crystallisation in a hydrate from the mass percentages
In this case you only have one variable to solve for ($x$) and you have one piece of information (Zn being 21.5% by mass) so you have enough information to solve the question.

The molecular mass of the hydrate can be expressed in terms of $x$:

$$\begin{align}
M(\ce{Zn(ClO3)2.xH2O}) &amp;= M(\ce{Zn}) + 2\cdot M(\ce{Cl}) + (6+x)\cdot M(\ce{O}) + 2x\cdot M(\ce{H}) \\
&amp;= \pu{65.38 g mol-1} + 2(\pu{35.45 g mol-1}) + (6+x)(\pu{16.00 g mol-1}) + \,\\
&amp;\qquad 2x(\pu{1.01 g mol-1}) \\
&amp;= (232.28 + 18.02x)~\pu{g mol-1}
\end{align}$$

Now

$$\begin{align}
\frac{M(\ce{Zn})}{M(\ce{Zn(ClO3)2.xH2O})} = \frac{\pu{65.38 g mol-1}}{(232.28 + 18.02x)~\pu{g mol-1}} &amp;= 0.215 \\
65.38 &amp;= 0.215(232.28 + 18.02x) \\
x &amp;= 3.99 \approx 4
\end{align}$$