A current of 1.93 A is passed through an aqueous solution of silver...

The total charge $Q$ passed through the solution is given by $Q = I \times t$.

Substituting the given values:
$Q = 1.93 \text{ A} \times 500 \text{ s} = 965 \text{ C}$

The reduction reaction at the cathode is:
$\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag}$

This indicates that $1 \text{ mole}$ of electrons ($96500 \text{ C}$) deposits $1 \text{ mole}$ of silver ($108 \text{ g}$).

Using Faraday's first law of electrolysis, the mass $m$ of silver deposited is:
$m = \dfrac{M}{nF} \times Q$

Here, $n = 1$ for $\text{Ag}^+$. Substituting the values:
$m = \dfrac{108}{1 \times 96500} \times 965$

$m = \dfrac{108}{100} = 1.08 \text{ g}$

Answer: $1.08 \text{ g}$