Solve this using method use in understanding chemistry textbook

$$t = 4 \text{ hours} \times \frac{60 \text{ minutes}}{1 \text{ hour}} \times \frac{60 \text{ seconds}}{1 \text{ minute}}$$ $$t = 14400 \text{ s}$$

The formula for charge is $Q = I \times t$.

$$Q = 3.0 \text{ A} \times 14400 \text{ s}$$ $$Q = 43200 \text{ C}$$

The formula for moles is $n = \frac{\text{mass}}{\text{molar mass}}$.

$$n_{Sn} = \frac{26.6 \text{ g}}{119 \text{ g/mol}}$$ $$n_{Sn} \approx 0.223529 \text{ mol}$$

One mole of electrons carries a charge equal to Faraday's constant ($F$).

$$n_e = \frac{Q}{F}$$ $$n_e = \frac{43200 \text{ C}}{96500 \text{ C/mol}}$$ $$n_e \approx 0.447668 \text{ mol}$$

From the half-reaction $\text{Sn}^{z+} + ze^- \to \text{Sn}(s)$, we know that $z$ moles of electrons are required to deposit 1 mole of tin. Therefore, the ratio of moles of electrons to moles of tin gives the oxidation state.

$$z = \frac{n_e}{n_{Sn}}$$ $$z = \frac{0.447668 \text{ mol}}{0.223529 \text{ mol}}$$ $$z \approx 2.0027$$

Rounding to the nearest whole number, as oxidation states are typically integers:

$$z = 2$$

The oxidation state of tin in the salt is $\text{+2}$.