Calculation of equivalent weight
I would express equivalent weight $M_\mathrm{eq}(\ce{E})$ of unknown element $\ce{E}$ using the number of equivalents $x$ and corresponding molecular weights $M$:

\begin{align}
M(\ce{ECl_x}) &amp;= x \cdot M_\mathrm{eq}(\ce{E}) + x \cdot M(\ce{Cl}) \tag{1} \\ \to \quad M_\mathrm{eq}(\ce{E}) &amp;= \frac{M(\ce{ECl_x}) - x \cdot M(\ce{Cl})}{x} \tag{1a}
\end{align}

To find $x$, let's apply to the balanced reaction, assuming it is complete and there is no chlorine left:

$$\ce{E (s) + x/2 Cl2 (g) -&gt; ECl_x (g)}$$

The amounts $n$ of gaseous products are bound by stoichiometry:

$$n(\ce{Cl2}) = \frac{x}{2}n(\ce{ECl_x}) \tag{2}$$

At constant temperature and pressure $n_i V_i = \mathrm{const}$. The problem says the volume of the system reduces by $1/3$, which means:

\begin{align}
V(\ce{Cl2}) &amp;= \frac{3}{2}V(\ce{ECl_x}) \tag{3} \\
\to \quad n(\ce{Cl2}) &amp;= \frac{3}{2}n(\ce{ECl_x}) \tag{3a}
\end{align}

Now, comparing/equating (2) and (3a) it's easy to see that $x = 3$:

$$\frac{x}{2} = \frac{3}{2} \quad \to \quad x = 3$$

You already correctly determined molecular weight from vapor density $\rho_\mathrm{v}$:

$$M(\ce{ECl_x}) = \rho_\mathrm{v} \cdot M(\ce{H2}) = 68.75 \cdot \pu{2.016 g mol-1} = \pu{138.60 g mol-1} \tag{4}$$

All that's left now is to do the math with (1a):

$$M_\mathrm{eq}(\ce{E}) = \frac{\pu{138.60 g mol-1} - 3 \cdot \pu{35.45 g mol-1}}{\pu{3 equiv}} = \pu{10.75 g mol-1 equiv-1}$$

I'm not sure this makes any sense from the chemistry prospective as this would suggest that element $\ce{E}$ is sulfur and so the gaseous product is $\ce{SCl3}$. I would've expected the compound to be something like $\ce{BCl3}$, but in this case the vapor density should've been $58.12$.