1. Write the electrode reaction

In $Fe_2(SO_4)_3$, iron is present as $Fe^{3+}$. To deposit iron metal:

$Fe^{3+} + 3e^- \rightarrow Fe$

So, 1 mole of Fe requires 3 moles of electrons.

2. Moles of iron deposited

Given mass of iron deposited = $0.3482\,g$

Molar mass of iron = $56\,g\,mol^{-1}$

$n(Fe)=\frac{0.3482}{56}=0.006217857\,mol$

3. Moles of electrons required

Since 1 mole Fe needs 3 moles electrons,

$n(e^-)=3\times 0.006217857=0.018653571\,mol$

4. Charge required

Using $1F = 96500\,C\,mol^{-1}$,

$Q = n(e^-)\times 96500$

$Q = 0.018653571\times 96500 = 1800.07\,C$

5. Use $Q = It$

Current $I = 1.5\,A = 1.5\,C\,s^{-1}$

$t = \frac{Q}{I} = \frac{1800.07}{1.5} = 1200.05\,s$

Convert to minutes:

$x = \frac{1200.05}{60} = 20.00\,\text{min}$

6. Nearest integer

$x = 20$

So, the required time is:

$\boxed{20}$