To solve this problem, we need to calculate the amount of Zn²⁺ ions that get deposited during the electrolysis and then use this information to find the final molarity of Zn²⁺ ions in the solution.

First, let's calculate the total charge passed through the cell:

$$ Q = I \times t $$

$$ Q = 1.70 \, \text{A} \times 230 \, \text{s} $$

$$ Q = 391 \, \text{C} $$

Next, we consider the current efficiency, which is 90%. Therefore, the effective charge used for the deposition of zinc is:

$$ Q_{\text{effective}} = 0.90 \times 391 \, \text{C} $$

$$ Q_{\text{effective}} = 351.9 \, \text{C} $$

Now, we use Faraday's laws of electrolysis to find the amount of zinc deposited. The molar mass of zinc (Zn) is 65.38 g/mol, and the charge per mole of Zn²⁺ (as it releases two electrons per ion) is:

$$ n_{\text{electrons}} = 2 \times F $$

$$ n_{\text{electrons}} = 2 \times 96485 \, \text{C/mol} $$

$$ n_{\text{electrons}} = 192970 \, \text{C/mol} $$

Using the effective charge, the moles of Zn deposited can be calculated as:

$$ \text{moles of Zn} = \frac{Q_{\text{effective}}}{n_{\text{electrons}}} $$

$$ \text{moles of Zn} = \frac{351.9 \, \text{C}}{192970 \, \text{C/mol}} $$

$$ \text{moles of Zn} = 0.001824 \, \text{mol} $$

The initial moles of Zn²⁺ in the solution can be calculated from its initial concentration and volume:

$$ \text{moles of Zn}\,^{2+}_{\text{initial}} = M \times V $$

$$ \text{moles of Zn}\,^{2+}_{\text{initial}} = 0.160 \, \text{M} \times 0.300 \, \text{L} $$

$$ \text{moles of Zn}\,^{2+}_{\text{initial}} = 0.048 \, \text{mol} $$

After deposition, the moles of Zn²⁺ decreases by the moles of Zn deposited:

$$ \text{moles of Zn}\,^{2+}_{\text{final}} = \text{moles of Zn}\,^{2+}_{\text{initial}} - \text{moles of Zn} $$

$$ \text{moles of Zn}\,^{2+}_{\text{final}} = 0.048 \, \text{mol} - 0.001824 \, \text{mol} $$

$$ \text{moles of Zn}\,^{2+}_{\text{final}} = 0.046176 \, \text{mol} $$

Finally, the final concentration of Zn²⁺ ions can be calculated maintaining the same volume:

$$ C_{\text{final}} = \frac{\text{moles of Zn}\,^{2+}_{\text{final}}}{V} $$

$$ C_{\text{final}} = \frac{0.046176 \, \text{mol}}{0.300 \, \text{L}} $$

$$ C_{\text{final}} = 0.1539 \, \text{M} $$

Thus, the final molarity of Zn²⁺ in the solution after 230 seconds of electrolysis with a current of 1.70 A and a current efficiency of 90% is approximately 0.154 M.