Equivalent weight of the anhydrous acid
Let the anhydrous acid be $\ce{HA}$ and assume its Molar Mass to be $X$. There is no a priori reason to assume $\ce{A}$ is a mere element.

The copper salt in question is $\ce{CuA2.3H2O}$. 

Its Molar Mass is:

$$63.55+2\times(X-1)+3\times 18=115.55+2X$$

Assume reasonably and without losses, that:

$$\ce{CuA2.3H2O(s) -&gt; CuO(s)}$$

So that $1\ \mathrm{mol}$ of the hydrated salt decomposes to $1\ \mathrm{mol}$ of cupric oxide.

The Molar Mass of $\ce{CuO}$ is $79.545$, so $0.3306\ \mathrm{g}$ of it is:

$$\frac{0.3306}{79.545}=4.156\times 10^{-3}\ \mathrm {mol}$$

So now we know that $1\ \mathrm{g}$ of the hydrated salt is $4.156\times 10^{-3}\ \mathrm {mol}$, so its Molar Mass is:

$$\frac{1}{4.156\times 10^{-3}}=240.6\ \mathrm{g/mol}$$

And with the above:

$$115.55+2X=240.6$$

Solve for $X$.

As the acid is monobasic, its Molar Mass is also its Equivalent Mass.