The volume of nickel plated is surface area times the thickness. That's $2m^2 \times 0.03 \ cm $ We'll use cubic cm as the units so it's $2m \times 1m \times 0.03 \ cm $ = $200 cm \times 100 cm \times 0.03 \ cm= 600 \ cm^3 $

This helps to find the mass as the density is given as $8.9 g/cm^3$. The mass = density $\times$ volume = $ 8.9 \ g/cm^3 \times 600 \ cm^3 = 5.34 \ kg$

Convert this to moles of Nickel. $\displaystyle \frac{5.34 \times 10^3 \ g}{58.71 \ g/mol} = 90.96 \ mol $ We'll come back to this.

Now for the current, $0.3 A/cm^2$ applied to a 5 cm square plate, meaning that the total current on the plate at a given time is $0.3 A \times 25 \ cm^2 = 7.5 \ A$ or $7.5 \ C/s$. However, only a fraction of that actually reduces the nickel (75%) so $5.625 \ A$

The voltage is then $ V = IR = 5.625 \times 0.4 $

The time calculation involves the actual nickel. $I = \displaystyle \frac{q}{t}$
Where q = total charge

We have 90.96 mols of Nickel reduced.

$Ni^{2+} + 2e^-$ -> $Ni$

That's 2 moles of electrons per mole of Nickel. So we need $181.92$ mol of electrons. The charge per mol of electrons is Faraday's constant of $96485.3 \ C/mol$ The total charge is just $181.92 mol \times 96485.3 \ C/mol$

Then we rearrange the formula of current for time $\displaystyle t = \frac{q}{I} = \frac{181.92 mol \times 96485.3 \ C/mol}{5.625 C/sec} $ $time = 3120463 \ seconds$ or $866 \ hours$

That's sort of the basic idea I'd imagine. I may have made a mistake somewhere in between. The time though while seems long is in line with this: https://sciencing.com/calculate-electroplating-7597391.html

There, they plated just one mole of Cu and with a higher current. That took 2 hours. We have a lot more of Ni and smaller current.

Then $Energy = power \times time = VIt$ and you can calculate the cost from there.

The volume of nickel plated is surface area times the thickness. That's $2m^2 \times \pu{0.03 cm} $.

We'll use cubic cm as the units so it is: $2m \times 1m \times \pu{0.03 cm} = \pu{200 cm} \times \pu{100 cm} \times \pu{0.03 cm} = \pu{600 cm3} $

This helps to find the mass as the density is given as $\pu{8.9 g/cm3}$.

The mass = density $\times$ volume = $ \pu{8.9 g/cm3} \times \pu{600 cm3} = \pu{5.34 kg}$

Convert this to moles of Nickel: $\displaystyle \frac{5.34 \times \pu{10^3 g}}{\pu{58.71 g/mol}} = \pu{90.96 mol} $. We'll come back to this.

Now for the current, $\pu{0.3 A/cm2}$ applied to a $5 \times \pu{5 cm}$ square plate, meaning that the total current on the plate at a given time is $\pu{0.3 A} \times \pu{25 cm2} = \pu{7.5 A}$ or $\pu{7.5 C/s}$. However, only a fraction of that actually reduces the nickel ($75\%$) so $\pu{5.625 A}$.

The voltage is then $ V = IR = 5.625 \times 0.4 $

The time calculation involves the actual nickel. $I = \displaystyle \frac{q}{t}$
Where $q =$ total charge

We have $\pu{90.96 mol}$ of Nickel reduced.

$\ce{Ni^{2+} + 2e^- -> Ni}$

That's $\pu{2 mol}$ of electrons per mole of Nickel. So we need $\pu{181.92 mol}$ of electrons.

The charge per mol of electrons is Faraday's constant of $\pu{96485.3 C/mol}$ The total charge is just $\pu{181.92 mol} \times \pu{96485.3 C/mol}$.

Then we rearrange the formula of current for time $\displaystyle t = \frac{q}{I} = \frac{\pu{181.92 mol} \times \pu{96485.3 C/mol}}{\pu{5.625 C/sec}}$ $\text{time} = \pu{3120463 seconds}$ or $\pu{866 hours}$.

That's sort of the basic idea I'd imagine. I may have made a mistake somewhere in between. The time though while seems long is in line with this: https://sciencing.com/calculate-electroplating-7597391.html

There, they plated just one mole of Cu and with a higher current. That took 2 hours. We have a lot more of $\ce{Ni}$ and smaller current.

Then $\text{Energy} = \text{power} \times \text{time} = VIt$ and you can calculate the cost from there.