I said answer want in English explanation

$$Q = I \times t$$

Where $I$ is the current and $t$ is the time.

$$Q = 1.5 \text{ A} \times 600 \text{ s} = 900 \text{ C}$$

$$\text{Cu}^{2+}(aq) + 2e^- \to \text{Cu}(s)$$

This reaction shows that 2 moles of electrons are required to deposit 1 mole of copper (Cu).

Faraday's constant ($F$) is $96485 \text{ C/mol e}^-$.
Moles of electrons ($n_e$) = $\frac{Q}{F}$

$$n_e = \frac{900 \text{ C}}{96485 \text{ C/mol e}^-} \approx 0.009328 \text{ mol e}^-$$

Since 2 moles of electrons deposit 1 mole of copper, the moles of copper deposited ($n_{\text{Cu}}$) will be:

$$n_{\text{Cu}} = \frac{n_e}{2} = \frac{0.009328 \text{ mol e}^-}{2} \approx 0.004664 \text{ mol}$$

The molar mass of copper is $63.5 \text{ g/mol}$.
Mass ($m$) = Moles $\times$ Molar mass

$$m = 0.004664 \text{ mol} \times 63.5 \text{ g/mol} \approx 0.2964 \text{ g}$$

The mass of copper deposited at the cathode is:

$$\text{0.2964 g}$$